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Probability

  • In the lesson on Counting, we defined the term event to simply mean something happening. For example, if we flipped a coin, we might name an event E that is the event of the coin coming up heads. Of course, there might be possibilities other than the event occurring. We call the set of all possible outcomes the sample space. If we want, we can denote the sample space with a symbol, such as S. In the coin example, there are two possible outcomes in the sample space S: heads and tails.
  • Let E be an event and S be the corresponding sample space. Let n(E) denote the number of ways E can occur and n(S) denote the total possible number of outcomes. Then if all the possible outcomes in S are equally likely, the probability of event E occurring (denoted P(E)) is
    P(E) = n(E)

    n(S)


     
    .
    Equivalently, using words,
    Probability of event = # of ways event can occur

    # of possible outcomes


     
    .
  • Notice that in the previous definition, it was assumed that all the possible outcomes were "equally likely". If this isn't true, the above does not work. Happily, we are almost certainly not going to see any problems that don't involve equal likelihood of all the outcomes. It might be described in the problem as `fair', `random', or something else, but we can almost always assume that all possible outcomes are equally likely at this level in math.
  • We can represent probability as a fraction ([1/2]), a decimal (0.5), or a percentage (50%). In any case, a probability is always between 0 and 1, inclusive. The larger the value, the more likely. We can also interpret probability as the ratio of the event happening over a large number of attempts. For example, if we flip a coin a million times, we can expect about half of the flips to come out heads. (P(Eheads) = [1/2])
  • Given two mutually exclusive events A and B, the probability of either one (or both) occurring (A∪B) is given by
    P(A ∪B) = P(A) + P(B) .
    If the events are not mutually exclusive, we have to take the overlap into account. We can represent where A and B overlap with A ∩B. Then
    P(A∪B)  = P(A) + P(B) − P(A∩B).
  • If we have some event E, we can talk about the event of E not occurring. We call this the complement of E, denoting it as Ec. [Other textbooks/teachers might denote it E or E′.] The probability of an event's complement occurring is
    P(Ec) = 1 − P(E).
  • Two events are independent if they are separate events and the outcome of either one does not affect the other. Given two independent events A and B, the probability of both events occurring is
    P(A and B)  = P(A) ·P(B).
  • If the events are not independent (the outcome of one does affect the other) and we want to find the chance of them both occurring, we need the idea of conditional probability. We denote the conditional probability of B occurring if A does occur as P(B | A). (We can interpret this as the probability of B happening if we are guaranteed that A will happen.) Then, given two events A and B, where the outcome of A affects the outcome of B, the probability of both events occurring is given by
    P(A and B)  = P(A) ·P(B | A).

Probability

Given a fair, six-sided die, what is the chance of rolling a 1, 5, or 6 on it?
  • A fair die is one where each side coming up is equally likely. When all the ways that something can occur are equally likely, the probability of something happening is quite simple:
    Probability of event = # of ways event can occur

    # of possible outcomes
  • For this problem, there are three different ways the event we are interested in can occur: we can roll a 1, we can roll a 5, or we can roll a 6.
    # of ways event can occur     =     3
    Since the die has six sides, there are a total of 6 different possible outcomes when we roll the die.
    # of possible outcomes     =     6
  • To find the probability of the event occurring, simply divide the number of ways the event we are interested in can happen by the total number of possible outcomes:
    P     =     3

    6
[1/2]
Given a standard deck of 52 playing cards, what is the probability of drawing a card at random from the deck and winding up with a 7, 8, or 9?
  • If you don't know how a standard deck of 52 playing cards is made up, start off by doing a little research. You need to know what you're working with before you can solve a problem. Looking it up, we find that a standard deck of cards has four suits (each with one of two colors): `Spades' (black), `Hearts' (red), `Clubs' (black), `Diamonds' (red). In each of these suits, there are 13 different cards. The cards are `Ace', the numbers 2-10, `Jack', `Queen', and `King'. Finally, the last three cards (J, Q, K) are sometimes called `Face' cards because they have a picture of a person on them.
  • Since we're told that we draw the card randomly from the deck, we know that we are equally likely to pull out any card, so we have
    Probability of event = # of ways event can occur

    # of possible outcomes
  • Counting up the number of 7's, 8's, and 9's in the deck, we see that they will appear once in every suit. Since there are four suits, that means we have 3·4=12 of them in the deck. The deck has a total of 52 cards, so the probability is
    P     =     12

    52
        =     3

    13
    [Notice that we could also get directly to the simplified fraction of [3/13] by the following logic: when we pull a card from the deck, it must be one of the suits. Thus we only need to know the chance of getting a 7, 8, or 9 within a single suit. Since there are 13 cards in each suit, we have [3/13].]
[3/13]
A group of eight people is going to be randomly seated around a circular table, where all the seats are evenly arranged. Within that group of eight people are two friends. What is the chance that the two friends will be seated next to one another?
  • Begin by visualizing the problem. There is a round table with eight chairs sitting around it. If you have difficulty understanding it, you might find it useful to draw a quick sketch: perhaps a circle with eight little boxes around it. Eight people will randomly sit down, and we're interested in the chance of two specific people sitting next to each other.
  • We can approach the problem like this: one of the two friends will be seated somewhere. We don't know where, but the first friend will have to be seated at one of the eight seats-this is guaranteed. From there, how many ways can the second friend be seated next to the first friend? How many ways could the second friend be seated total?
  • Remember, since it's a circular table, the second friend could be seated to the right or to the left of the friend. Thus there are 2 possible ways they could be seated next to each other. After the first friend has been seated, there are only seven seats left at the table (8−1), so there are a total of 7 possible ways for the second friend to be seated. Since the seating is random, we have a probability of [2/7].
[2/7]
A fair, six-sided die is rolled three times in succession. What is the probability that the first roll will be a 4, 5, or 6, the second roll will be a 5 or 6, and the third roll will be a 6?
  • Begin by noticing that each roll of the die is unaffected by the other rolls. The probability of the first roll coming up 4, 5, or 6 is unaffected by what happens in the other rolls, and similarly for the second and third rolls being unaffected by the others. This means that we have independent events: we can find the probability of all the events occurring by multiplying each of the sub-events together.
  • The probability of the first roll coming up 4, 5, or 6 is [3/6] (3 ways it can happen, 6 possible outcomes). The probability for the second roll coming up 5 or 6 is [2/6] (2 ways it can happen, 6 possible outcomes). The probability for the third roll coming up 6 is [1/6] (1 way it can happen, 6 possible outcomes).
    Pfirst = 3

    6
    ,        Psecond = 2

    6
    ,        Pthird = 1

    6
  • Since each of the three rolls must occur for the event we are interested in (the event is the three rolls each coming out as specified in the problem), we can multiply them all together to find the probability of the event:
    P     =     Pfirst ·Psecond ·Pthird     =     3

    6
    · 2

    6
    · 1

    6
        =     1

    2
    · 1

    3
    · 1

    6
        =     1

    36
[1/36]
At a certain school, on any given day the probability that a random student does not eat anything for breakfast is 23%. If the school has a total of 1200 students, how many students eat breakfast on any given day?
  • Begin by noticing that `not eating anything for breakfast' and `eating breakfast' are opposite conditions. If one occurs, the other must not, and vice-versa. In "math-speak", these are called complementary events: if something happens, it must be in one of the events, and it cannot be in both of them.
  • The probability of an event's complement (its "opposite") occurring is
    1−P,
    where P is the probability of the event occurring. For this problem, we were given the probability as P = 23%, so we must first convert it to decimal form: P = 0.23. Once we have that, we can find the probability of the complement:
    1− 0.23     =     0.77
    [Notice that we could also do this with percentages: 100% − 23% = 77%. Converting to decimal format is convenient for the next step, though.]
  • Now that we know the probability of a student eating breakfast is 0.77, we can multiply that by the total number of students to find what portion of them will fulfill that event-that is, eat breakfast:
    0.77·1200     =     924
924
You have lots of fair, six-sided dice in front of you. What is the minimum number of dice you must roll at the same time so that the probability of at least one of them coming up with a 5 is at least 50%?
  • Begin by understanding the question. You can roll some number of dice (you choose how many you roll), then you look at the results of those dice. Your job is to figure out how many dice need to be rolled for the chance of at least one die coming up as a 5 to be greater than or equal to 50%.
  • Before we solve the problem, let's look at a common mistake people make when attempting to solve this kind of problem: lots of people assume that since the chance of one die coming up as a 5 is [1/6], if we roll six dice, we will be guaranteed a 5. Since that is 100%, half as many dice would produce 50%, so three dice is the required number. The above logic is faulty for a couple of reasons, but the simplest one is that you can never guarantee a certain number will be rolled on a fair die. If you roll a hundred dice, there is always the chance that they could all come out as 1's, or anything else. The chance may be tiny at that point, but it is impossible to be completely certain. That 100% probability the "logic" was built on is impossible to achieve.
  • Instead, a good way is to figure out the chance of a 5 not being rolled. Once we know the chance of that happening, we can figure out the complement: the chance of one or more 5's being rolled. For example, at one die, the chance of a 5 not being rolled is [5/6]. Thus the chance of a 5 coming up is
    One die:        1 − 5

    6
        =     1

    6
  • Work through this with increasing number of dice until you find the first one that a probability of 50% occurs at. For two dice, the chance of a 5 not being rolled on either die is [5/6] ·[5/6], so the probability of a 5 coming up somewhere on two dice is
    Two dice:        1 −
    5

    6
    · 5

    6

        ≈     0.306
    For three dice, the chance of a 5 not coming up is [5/6] ·[5/6] ·[5/6], so the probability of a 5 coming up somewhere on three dice is
    Three dice:        1 −
    5

    6
    · 5

    6
    · 5

    6

        ≈     0.421
    For four dice, the chance of a 5 not coming up is [5/6] ·[5/6] ·[5/6] ·[5/6], so the probability of a 5 coming up somewhere on four dice is
    Four dice:        1 −
    5

    6
    · 5

    6
    · 5

    6
    · 5

    6

        ≈     0.518
    Thus the minimum number of dice to guarantee a 50% or better chance is four dice.
4
Given a standard deck of 52 playing cards, what is the chance of randomly drawing three cards and them all being face cards?
  • If you don't know how a standard deck of 52 playing cards is made up, start off by doing a little research. You need to know what you're working with before you can solve a problem. Looking it up, we find that a standard deck of cards has four suits (each with one of two colors): `Spades' (black), `Hearts' (red), `Clubs' (black), `Diamonds' (red). In each of these suits, there are 13 different cards. The cards are `Ace', the numbers 2-10, `Jack', `Queen', and `King'. Finally, the last three cards (J, Q, K) are sometimes called `Face' cards because they have a picture of a person on them. Since there are three face cards for each suit, and there are four suits in a deck, a 52 card deck begins with 3·4 = 12 face cards.
  • Notice that, unlike previous problems, the card that gets pulled first affects the probability of later cards being pulled. If a face card is pulled, there is less chance of pulling a face card later. If a face card is not pulled, there is more of a chance of pulling a face card later. To deal with the fact that each pull affects the subsequent ones, we can use the idea of conditional probability. If we know the chance of a first event occurring, and we know the chance of a second event occurring assuming the first event occurs, we can multiply them together to find the chance of both events occurring. For this specific problem, that means we can find the chance of the first pull being a face, then the chance of the second (assuming the first pull was a face), then the chance of the third (assuming the first two pulls were faces), then finally multiply them all together.
  • Since a 52 card deck has 12 face cards to begin with, the chance of the first face being pulled is
    First face:        12

    52
    Now we move on to the next card. At this point, the deck has one less card and one less face, so there are 51 cards with 11 faces shuffled in:
    Second face:        11

    51
    Finally we have the third card. At this point, the deck is two cards down and two faces down, so there are 50 cards with 10 faces:
    Third face:        10

    50
    To find the probability of the event occurring where each pull happens, we multiply them all together:
    P     =     12

    52
    · 11

    51
    · 10

    50
        =     11

    1105
        ≈     0.009  955
[12/52] ·[11/51] ·[10/50]     =     [11/1105]     ≈     0.009  955
In a sack of marbles, there are 8 green marbles, 12 blue marbles, 15 red marbles, and 5 yellow marbles for a total of 40 marbles in the sack. If you reach into the sack and draw out three marbles at random, what is the chance that you will draw precisely 1 green marble and 2 red marbles?
  • Notice that drawing a given marble affects the chances of drawing other marbles. This means that if we want to approach this problem in a step-by-step manner (first marble, second marble, third marble), we need to use conditional probability. However, unlike the previous problem, there are multiple ways for the event we are interested in to occur. The event could happen in a variety of different orders, shown below (where G means a green marble is pulled and R means a red marble is pulled):
    Possible orders:       GRR,       RGR,        RRG
  • Since the event can occur using any of these three different ways, we will sum the probabilities for each of the orders to find the total probability of the event. Working through each order with conditional probability, we have
    GRR        ⇒        8

    40
    · 15

    39
    · 14

    38

    RGR        ⇒        15

    40
    · 8

    39
    · 14

    38

    RRG        ⇒        15

    40
    · 14

    39
    · 8

    38
  • Thus, since the event can happen by following any of these three orders, we add them all together to find the probability of the event happening by any of the orders:

    8

    40
    · 15

    39
    · 14

    38

    +
    15

    40
    · 8

    39
    · 14

    38

    +
    15

    40
    · 14

    39
    · 8

    38

    Simplify:
    15 ·14 ·8

    40 ·39 ·38
    + 15 ·14 ·8

    40 ·39 ·38
    + 15 ·14 ·8

    40 ·39 ·38
        =     3 ·
    15 ·14 ·8

    40 ·39 ·38

        ≈     0.085
  • Alternatively, there is another way to do this that does not involve conditional probability and is arguably simpler. Remember, since the drawing is random, we have
    Probability of event = # of ways event can occur

    # of possible outcomes
    Using combinations from the previous lesson, it's easy to see how many total possible outcomes there are: we're choosing three marbles from a bag of 40, so it's 40C3. To figure out the number of ways that 1 green and 2 reds can be pulled, look at the number of ways each marble can be pulled from its group: there are 8C1 ways to pull the green and 15C2 ways to pull the two reds. If we want to know all the ways we can do both, we multiply them because each green still allows the pulling of reds, and each red still allows the pulling of a green. Thus the total number of ways the event can happen is (8C1) ·(15C2). This means the probability of the event is
    (8C1) ·(15C2)

    40C3
        =    
    8!

    1!·7!
    · 15!

    2! ·13!

    40!

    3!·37!
    Simplify:
    8!

    1!·7!
    · 15!

    2! ·13!

    40!

    3!·37!
        =    
    8

    1
    · 15 ·14

    2

    40·39 ·38

    3 ·2
        =     8 ·15 ·7

    40 ·39 ·38
    ·3 ·2     =     3 ·
    15 ·14 ·8

    40 ·39 ·38

        ≈     0.085
    Which is exactly what we got doing it with conditional probability, so the two methods check each other. Great!
3 ·( [(15 ·14 ·8)/(40 ·39 ·38)] )     ≈     0.085
At a popular game show, four contestants are randomly selected from the audience. At today's filming for the show, there are 47 students from a nearby college. If the total number of people in the audience is 400, what is the chance that precisely two students from the college will be picked to be contestants?
  • Note: To do this problem, we will use the idea of a combination from the previous lesson. If you are not familiar with this idea, you will need to watch the part of the previous lesson dealing with combinations before the below makes sense.
  • Begin by making sense of the problem: there is an audience comprised of two groups-college and non-college. There are a total of 400 people, with 47 of those being in the college group. From the entire audience of 400, four different people are selected. Our job is to figure out what the probability is of precisely 2 college and 2 non-college being picked together.
  • We might consider doing this problem by creating a lengthy tree diagram, where each branch represents one of the possible events (college or non-college picking). We can calculate the probability of a given path occurring by multiplying together all the probabilities for the branches in that path. Then we can find the total probability of our event (2 college and 2 non-college) occurring by adding up all those paths. This method will work. It is also extremely difficult and time-consuming to do the above. Instead, there is a much easier way. Remember, since the selection is random, the probability of the event happening is quite simple:
    Probability of event = # of ways event can occur

    # of possible outcomes
  • This means all we need to do is compute the number of ways the event can occur and the number of all possible outcomes. To figure out the ways the event can occur, notice that we're interested in 2 college and 2 non-college being picked. Since we have 47 college students total, all the ways that 2 college students can be chosen is 47C2. We can work things out similarly for all the ways 2 non-college can be picked: there are 400−47=353 non-college in the audience, so the number of possible ways is 353C2. Thus, the number of possible ways both things can happen is the product:
    # of ways to pick 2 college and 2 noncollege:        (47C2) ·(353C2)
    The total number of possible outcomes is much simpler: it's just all the ways that 4 people can be selected from an audience of 400:
    # of ways to pick 4 people from audience:        400C4
  • To find the probability, we divide the ways the event can happen by all the possible outcomes, then simplify:
    (47C2) ·(353C2)

    400C4
        =    
    47!

    2!·45!
    · 353!

    2!·351!

    400!

    4!·396!
        =    
    47·46

    2
    · 353 ·352

    2

    400 ·399 ·398 ·397

    4 ·3 ·2
    To make it easier to see, we can multiply the numerator by the denominator's reciprocal:
    47 ·46 ·353 ·352

    2 ·2
    · 4 ·3 ·2

    400 ·399 ·398 ·397
        =     47 ·46 ·353 ·352 ·6

    400 ·399 ·398 ·397
    Finally, use a calculator to simplify and get an approximate decimal:
    16 790 092

    262 684 975
        ≈     0.0639
[(16 790 092)/(262 684 975)]     ≈     0.0639
If you randomly draw five cards from a standard deck of 52 playing cards, what is the percentage chance that you will draw a full house?
  • Note: To do this problem, we will use the idea of a combination from the previous lesson. If you are not familiar with this idea, you will need to watch the part of the previous lesson dealing with combinations before the below makes sense.
  • If you don't know what a full house is or how a standard deck of 52 playing cards is made up, start off by doing a little research. You need to know what you're working with before you can solve a problem. Looking up full house, we find that it is a set of five cards where 3 of the cards all match each other and the remaining 2 also match. Notice that a full house has nothing to do with the suits of the cards: they merely need to match in terms of being the same number/face/ace. Thus a full house contains one three-of-a-kind and another pair. Some examples: 777KK,  AAA33,  88855, etc. From previous problems, we've learned that a standard deck of cards has four suits. In each of these suits, there are 13 different cards. The cards are `Ace', the numbers 2-10, `Jack', `Queen', and `King'.
  • We might consider doing this problem by creating a lengthy tree diagram, where each branch represents one of the possible events (which card is picked next). We can calculate the probability of a given path occurring by multiplying together all the probabilities for the branches in that path. Then we can find the total probability of our event (full house) occurring by adding up all those paths. This method will work. It is also extremely difficult and time-consuming to do the above. Instead, there is a much easier way. Remember, since the selection is random, the probability of the event happening is quite simple:
    Probability of event = # of ways event can occur

    # of possible outcomes
  • This means all we need to do is compute the number of ways a full house can occur along with the total number of ways five cards can be drawn from 52. First off, notice that the order of the cards is unimportant: you still have a full house whether you draw in the order of 33322 or 2 33 2 3 or anything else. This means we will be using combinations, not permutations. This problem is not about ordering, it is about choosing. To compute the number of ways to draw three-of-a-kind and a pair, let's start with the three-of-a-kind first. We can do this with any of the 13 ranks of card, and once we choose a rank (A,2,3,…,Q,K), we then need to pull three of them out of the four total. Thus, for the three of a kind, there are 13 ·(4C3). For the pair, we can do it with any of the remaining 12 ranks, and we pull out two of them, so 12 ·(4C2). To find all the ways to do both of these, we multiply them together:
    # of ways to draw full house:       
    13 ·(4C3)
    ·
    12 ·(4C2)
    To find the total number of ways we can draw five cards, that's much simpler: all the ways that five objects can be chosen from 52.
    # of ways to draw 5 cards:        52C5
  • To find the probability, divide the ways the event can happen by all the possible outcomes:
    13 ·12 ·(4C3) ·(4C2)

    52C5
        =    
    13 ·12 · 4!

    3!·1!
    · 4!

    2!·2!

    52!

    5!·47!
    Simplify a bit:
    13 ·12 ·4 · 24

    4

    52 ·51 ·50 ·49 ·48

    120
        =     13·12 ·4 ·6 ·120

    52 ·51 ·50 ·49 ·48
        ≈     0.00144
    Finally, the problem asked for a percentage chance, not a decimal, so convert:
    0.00144     =     0.144%
0.144%

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Probability

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:06
  • Definition: Sample Space 1:18
    • Event = Something Happening
    • Sample Space
  • Probability of an Event 2:12
    • Let E Be An Event and S Be The Corresponding Sample Space
  • 'Equally Likely' Is Important 3:52
    • Fair and Random
  • Interpreting Probability 6:34
    • How Can We Interpret This Value?
    • We Can Represent Probability As a Fraction, a Decimal, Or a Percentage
  • One of Multiple Events Occurring 9:52
    • Mutually Exclusive Events
    • What If The Events Are Not Mutually Exclusive?
    • Taking the Possibility of Overlap Into Account
  • An Event Not Occurring 17:14
    • Complement of E
  • Independent Events 19:36
    • Independent
  • Conditional Events 21:28
    • What Is The Events Are Not Independent Though?
    • Conditional Probability
    • Conditional Events, cont.
  • Example 1 25:27
  • Example 2 27:09
  • Example 3 28:57
  • Example 4 30:51
  • Example 5 34:15

Transcription: Probability

Hi--welcome back to Educator.com.0000

Today, we are going to talk about probability.0002

Consider if we wanted to know the chance of rain today, the odds of winning a bet, or how likely a medicine is to cure a disease.0004

In all of these cases, we would be asking the probability of something happening.0010

The study of probability and chance is a major area of mathematics.0014

It has applications throughout business, science, politics, medicine, and many other fields.0018

Being able to know how probable an event is, how likely something is to happen, is extremely important for a huge number of things.0022

In this lesson, we will go over some of the basic concepts of probability.0029

Still, even though they are basic concepts, it is going to be a really interesting amount of stuff that is going to let us see some really interesting results.0032

There is a lot of stuff here, even at the basic level.0039

Furthermore, basic probability questions pop up a lot on standardized tests, like the SAT and other standardized tests.0042

So, if you are planning on taking any of those in the near future, this is especially useful,0049

because you are almost certainly going to see some questions that are exactly like what we are working on here, but probably even easier.0052

And while it is not absolutely necessary to have watched the previous two lessons before watching this one,0057

we are going to draw very heavily on the previous two lessons in this lesson.0066

So, I would really recommend watching those first, if you haven't watched them.0069

It is not absolutely necessary, but it will make things a little clearer.0072

All right, let's go: the first thing that we want to define is the idea of a sample space.0075

In the lesson on counting, we defined the term event to simply mean something happening.0079

For example, if we had a coin, and we flipped it, we might name an event E that is the event of the coin coming up heads.0085

Of course, there might be possibilities other than the event occurring.0092

It might come up something other than heads.0095

We call the set of all possible outcomes, everything that could happen when we do something, the sample space.0097

If we want, we can denote the sample space with a symbol, such as S.0104

In the example above, there are two possible outcomes for the sample space.0108

1) It comes up heads, or 2) it comes up tails.0111

There are two different possibilities, because there are two different sides to the coin.0115

Two sides to the coin means two things in our sample space.0120

The event is simply heads coming up; the sample space is everything that could occur, heads and tails.0123

Probability of an event: let E be an event, and S be the corresponding sample space.0131

Let n(E) denote the number of ways that E can occur, and n(S) denote the total possible number of outcomes.0136

Then, if all of the possible outcomes in S are equally likely, the probability of event E occurring,0143

denoted p(E), probability of E, is p(E) = n(E)/n(S).0149

Equivalently, just using words, that is: the probability of an event is equal to the number of ways the event can occur, divided by the number of possible outcomes.0155

For example, in a standard 52-card deck, there are 4 of each card in various suits.0167

If we draw a card at random from the deck, the chance of drawing an ace is...0172

The probability of the event of an ace coming up is: there are four ways that we can get an ace out of the deck.0177

There are four aces in there, so there are four ways that that can happen.0187

And there are a total of 52 cards in the deck, so there is a total possibility...the total number of possible outcomes0190

is pulling any of those 52; so it is 4/52, which simplifies to 1/13.0195

This idea is the basic idea of probability: being able to say the number of ways it could occur, divided by the number of all possible things that could happen.0201

That is pretty much the main idea; and if you take one idea away from this lesson, this is the one idea to take away.0210

And this is the sort of thing that you will end up seeing on any standardized test.0215

This idea right here is enough for any standardized test.0218

So, as long as you keep that one in your mind, you will be good for all of that.0220

We are going to get into slightly more interesting things as we keep going, but this is the basic, fundamental idea that you want to hold onto.0223

Equally likely is important: that is a really important phrase.0231

In that previous definition, it was assumed that all possible outcomes were equally likely.0234

This is a really important requirement for the way that we are going to look at probability.0243

Why? Well, let's consider the following scenario.0247

If you dig a hole in the ground, there is a possibility that you will find gold.0249

The sample space, then, has two possibilities: you find gold, or you don't find gold.0255

We have an event, which is finding gold; and our sample space is finding gold or not finding gold.0261

That means one thing for the event is divided by two things for the size of our sample space.0267

But clearly, if you dig some random hole, the chance of you finding gold is not 1/2.0272

Why? Because each chance is not equally likely.0278

Each outcome is not equally as likely as the other one.0282

Finding gold is not equal occurrence, equal probability, with not finding gold; they are not equally likely outcomes.0285

So, because they are not equally likely outcomes, we can't base it off of this idea of0292

number of ways of our event, divided by number of possible ways anything could happen.0295

All right, the method that we just talked about for that basic probability thing won't work if we don't have this "equally likely" thing.0301

This "equally likely" thing is a really important first requirement.0309

Now, happily, we are almost certainly never going to see anything that doesn't involve equal likelihood.0314

All of the problems that we are going to end up seeing are going to be equal likelihood problems.0320

We are going to know that all of the outcomes are equally likely.0325

They might describe it with words like fair, like "a fair die" or "a fair coin" or a "fair" random number--0328

something that implies that all of the possibilities are equally likely.0337

A fair die is one that is equally likely to come up 1, 2, 3, 4, 5, 6.0342

All of its outcomes are equally likely.0347

We can also use the word "random"; if something is selected randomly, that is implying that, out of the selection, it was equally likely amongst all of them.0349

Or some other way of saying this...but we can almost always assume that all of the possible outcomes are equally likely.0356

At the level of the problems that we are going to be working at, this is a pretty reasonable assumption to make:0364

that we can assume this "equally likely" thing at the level of the problems we are working at.0368

There are lots of really interesting problems you can work on, where this assumption won't end up being true.0373

But for the sort of thing you will be required to work on at this point,0378

you can almost always be certain that you will be allowed to assume0381

that everything is equally likely, unless they very explicitly tell you otherwise.0384

And that is not going to happen very often, if at all.0387

All right, interpreting probability: notice that the probability of an event, p(E) = n(E)/n(S), is always less than or equal to 1.0393

This is because n(E) ≤ n(S); n(E), the number of ways the event can occur,0401

is always less than or equal to the total number of things that can happen,0407

because all of the ways that the event can occur are all inside of the ways that anything could happen.0410

E is always contained within our sample space; the event is always contained within the sample space,0416

just like heads was contained inside of heads and tails for the sample space.0422

The event is always contained inside of the sample space; so the number of ways the event can happen is always going to be smaller,0427

or equal to, the number of things in the sample space, total.0432

OK, we have this number, then, that can be somewhere between 0 and 1.0436

The smallest E could be is 0 ways total; so we are somewhere between 0 and 1.0440

How do we interpret this value?0444

We interpret it like this, where we have this value that can range between 0 and 1.0446

And at 0, it is absolutely impossible for the thing to occur; and at 1, it is absolutely certain that the thing will occur.0450

It will definitely occur at a 1 probability, and it will never occur at a 0 probability.0459

So, as we end up going up the scale, as we go up from 0 to 1, it becomes more and more likely.0463

The closer you get to 1, the more likely it is.0470

And here in the middle, at 0.5, it is equally likely as unlikely.0472

On average, 1 out of 2 times it will end up happening.0476

We can represent probability as a fraction, as a decimal, and as a percentage.0481

Any of these are fine things to do; the important part is that probability is always between 0 and 1, inclusive,0487

because you can be 0, and you can be 1, as a probability,0494

although almost all of the ones we are going to deal with will be somewhere in between.0497

We can also interpret probability as the ratio of the event happening over a large number of attempts.0500

For example, if we flip a coin a million times, we can expect about half of the flips to come out heads,0506

because the probability of flipping a fair coin and having it come out heads is 1 out of 2, 1/2.0512

On the large scale, we know that about half of any large thing will end up coming out to be that.0518

Now, on the small scale, if I flip a coin twice, it wouldn't be that surprising for two of them to come out as tails,0523

even though it is a 1-out-of-2 chance for heads.0529

We could flip the coin three times, and it wouldn't be that surprising for it to come out as tails, tails, tails.0532

It is not super likely, but it is not that unreasonable.0537

Just because the probability is 1/2 for heads doesn't guarantee us that it is going to occur any time.0540

With probability, we don't have a guarantee of occurrence; we just have "likely" that it will occur at certain levels of likelihood.0546

We can only have certainty at a 1.0554

So, since it is a question of how likely it is, we won't have it be as likely to show up, unless we look at a larger sample space.0556

We look at a larger number of things that could happen.0565

If we look at it happening a million times, we can be almost sure to have half of them be heads,0570

because we have done it so many times that we start to see this happen more and more.0575

On a very small scale, though, we can't be certain that it will end up showing up.0579

We flip a coin three times; it might come up tails all three times, but that doesn't imply that the coin isn't fair.0583

It is just how random chance works.0588

All right, one of multiple events occurring; consider if we wanted to find the probability of rolling a fair 6-sided die and having it come up either 1 or 6.0591

Now, we could consider them as separate events;0600

so we would call them E1 and E6, the event of a 1 and the event of a 6.0601

We can talk about either E1 or E6 occurring with the notation E1 union E6.0606

That is a way of saying E1 or E6 or both.0613

What we are looking for is something that happens, which is inside of E1 or inside of E6.0616

We can be inside of either of them, so it is a union.0622

What we are curious to know here: we are looking for the probability of rolling a 1 or a 6.0625

We are looking for the probability of E1 union E6, the probability of E1 or E6 or both of them.0630

Notice that E1 and E6 are mutually exclusive events.0638

If one of them occurs, the other one cannot occur.0642

What this means is that, if we roll a 1, it is impossible to have rolled a 6 just then.0646

If we rolled a 6, it is impossible to have rolled a 1 just then.0650

There is no overlap between them; we can't be a 1 and a 6 simultaneously.0653

So, they are mutually exclusive events.0657

With this in mind, we see that the probability of E1 or E6 occurring, E1 union E6, just combines their probabilities.0659

The probability of E1 union E6 is equal to the probability of E1, plus the probability of E6.0667

Because we don't have to worry about them overlapping, it is just "did E1 happen?" and then we also could look at "did E6 happen?"0673

The probability of E1 is 1 out of 6; the probability of E6, rolling a 6, is 1 out of 6.0680

We add those two together, and we get 1/3 as the total probability for what we would have of rolling a 1 or a 6.0685

This idea works in general: given two mutually exclusive events, A and B,0694

the probability of either one, or both, occurring, A union B, is given by:0700

the probability of A union B is equal to the probability of A, plus the probability of B.0704

Given that two events can't both happen simultaneously, if we know that, if you are A, then you can't be B,0710

and if you are B, then you can't be A, then if we are looking for either one of them happening,0717

it is just going to be adding the two probabilities together.0722

This is another one of those basic ones that you end up seeing on tests and homework.0724

This is a good one to remember.0729

What if the events are not mutually exclusive--there is some overlap in the events, so that you could be in A and B at the same time?0732

For example, let's consider the probability of a fair die coming up strictly below 4 and/or (so it can also be) coming up even.0740

So, the die comes up below 4 (that is 1, 2, 3), or the die comes up even.0751

Notice: there is some overlap in these two events.0757

The die could come up as 2, which is below 4, and also which is even.0761

It is both of these at the same time; it is both of the things.0768

Since it is both of the things, there is overlap between being below 4 and being even.0772

To find the probability, we have to take this overlap into account.0777

In two events, A and B, we denote the overlap of both occurring at the same time as0781

A intersect B, that is to say, where A and B are happening at the same time.0786

So, where A intersects with B is their area of overlap, where both things are happening at the same time.0792

We take the possibility of overlap into account as follows: let A and B be two events.0800

The probability of A or B or both is given by probability of A union B, that is to say, A or B, or both occurring,0804

is equal to the probability of A, plus the probability of B, minus the probability of A intersect B.0813

That is, minus the probability of A and B occurring at the same time.0820

So, this is an interesting formula; but that one that we just talked about, where we assumed that they are mutually exclusive--0825

that one is much more likely to come up in homework and tests.0830

You might end up seeing the formula that we are working on right now.0834

But it is less likely; so if this one doesn't make quite as much sense to you, don't worry about it too much.0838

We are about to see a quick example, though, that will help explain it.0842

So, that might help cement it; but don't worry about it too much if this one doesn't make a lot of sense.0845

You are much less likely to see it than the previous one.0848

Our previous example, this example that we are looking for, 1, 2, or 3, or an even number, or both of them:0852

we can let E123, which is the event of rolling a number strictly below 4,0858

which is to say a 1, a 2, or a 3; and then Eeven, which is rolling an even number (you roll a 2, a 4, or a 6)...0863

now notice where E123 and Eeven overlap.0873

These two things overlap at E2, when we roll a 2.0878

So, if a 2 comes up on the die, it is below 4, and it is even.0884

So, E2 is where they overlap each other.0889

By the above formula, the probability of A union B, the probability of E123 union Eeven,0893

of having it be either below 4 or even or both, is equal to the probability of the die coming up as 123,0898

plus the probability of the die coming up even, minus the probability of the die coming up both of these0909

(the probability of it being 123 and even).0916

So, what is the probability of it coming up as 123?0920

Well, 1, 2, 3...that is 3 possibilities, divided by 6 total on the thing, so it is 3/6.0922

Plus coming up even: 2, 4, 6; that is 3 possibilities, divided by 6 total, so 3/6 as well.0928

Minus...and now we can swap, since we know that E2 is the same thing as E123 intersect Eeven.0935

So, it is the same thing as just asking what the probability is of rolling a 2.0940

We combine red and blue together, and they come to be a 1; 3/6 + 3/6 is 1.0944

And then minus...what is the probability of rolling a 2?0951

Well, that is 1 (rolling a 2), divided by 6; so we have 1/6; 1 - 1/6 comes out to be 5/6.0954

So, we have 5/6 as the chance of rolling a number that is below 4, or even, or both.0963

Now, 5/6 makes sense, because we can also just go through this and do this by hand.0970

1, 2, 3, 4, 5, 6: notice: if you are 1, 2, or 3, then these ones are good; if you are even: 2, 4, and 6--then those ones are good, as well.0974

So, the only one that fails to be 1, 2, 3, or even is the number 5; that means we have0993

1, 2, 3, 4...1 here; 2 here; 3 here; 4 here; 5 possibilities for this event to occur,0999

divided by a total of 6 possible outcomes; so 5 divided by 6 is the exact same thing, so this checks out and makes sense.1008

So, this formula here makes sense on the small scale, and we can also bring it to a much larger scale,1016

if we are working with a much more complicated problem, where we can't just do this by hand,1021

and we have to be able to understand the theory to be able to get an answer.1024

All right, what if an event doesn't occur?1029

If we have some event E, we can also talk about the event of E not occurring.1031

We have E that occurs when the event E occurs; but we can also talk about if E does not occur; let's make that an event.1035

E not occurring is now an event, as well; we call this the complement of E.1042

The complement of an event is that event not occurring, and we denote it with E with a little c in the top corner,1047

so E to the c, but not actually raising it like an exponent; it is not the same thing...E complement.1054

Other textbooks or teachers might denote this as E with a bar on top or E with a little tick mark.1060

It doesn't matter; any of them is fine; but I am going to use Ec.1066

For example, if E is the event of rolling an even number on a die, then Ec is the opposite event.1070

In this case, we have that E is an even number; so Ec would be the complement,1077

when E does not occur; so if E does not occur, what is the opposite to rolling an even number?1082

That is rolling an odd number; so if you roll an odd number on the die, you are in the complement of the event here.1089

The probability of an event's complement occurring is 1 - the probability of the original event.1096

So, the probability of an event's complement is equal to 1 minus the probability of the original event.1102

Why? Well, either E occurs, or it does not; and if it does not, then we know that Ec must occur.1109

So, Ec must occur if we end up having that E does not occur.1118

If E occurs, then E has occurred; if E does not occur, then Ec has occurred.1124

So, no matter what, we can be certain that one of them must occur.1130

One of these two things always has to happen.1134

We can't have something in the middle: either the event happens, or the event doesn't happen.1137

So, if the event happens, or it doesn't happen, well, either way, one of those two things happened.1141

So, we are certain that one of them will occur.1146

Since one of them always has to occur, that means that the total of their probabilities must be a 1, certainty.1150

The probability of an event's complement, plus the probability of the event, is equal to 1,1156

because one of those two things always has to happen.1160

So, that is why we end up having 1 minus the probability of the event give us the probability of the event not happening.1164

This is a useful idea in a lot of situations.1170

So, this is another useful one to remember.1172

Independent events: consider rolling a die and flipping a coin.1175

How can we find the probability of the die coming up as a 5 or a 6, and the coin coming up heads?1179

To do this, we must consider the probabilities of both events.1186

In the example of the above events, we say that they are independent events,1189

because rolling the die has no effect on flipping the coin, and flipping the coin has no effect on rolling the die.1193

They are separate events, and the outcome of one event does not affect the other.1198

We know that they are independent events in this case.1203

If they are independent events, given two independent events, A and B, the probability of both events occurring,1206

that is to say, the probability of A and B occurring, is equal to the probability of A, times the probability of B.1213

We multiply the probabilities of each of them on their own, and that gives us the probability of both of them occurring, if they are independent events.1222

In the above example, we would have: the probability of an event of a 5 or 6, and the event of a heads,1230

is equal to the probability of the event of a 5 or a 6, times the probability of the event of a heads.1239

Well, what is the chance of a 5 or a 6?1244

A 5 or a 6 is going to be 2 out of 6; and the probability of a heads is going to be 1 out of 2.1246

We multiply these together, and we get 1/6.1253

This right here, this idea of multiplying probabilities together, is the other really important idea for this lesson.1259

That very, very basic one of the number of ways that the event can happen, divided by total number of things that can occur--1266

that is the first really important idea in this lesson, and the second one is:1272

if you get independent events, and you want to figure out the chance of both of them occurring,1275

you just multiply the probabilities of each of them on their own.1279

That one also comes up on tests a lot; that is another very important idea to take away from this.1282

What if the events are not independent, though--what if we are looking at a situation1287

where they aren't actually going to not affect each other, where they can have an effect on each other?1290

The outcome of one will do something to the outcome of the other, or at least the chances of the outcome of the other.1296

For example, consider of we draw two cards at random from a deck of 52 cards.1301

What is the probability of a pair of aces--that is, both cards coming out as aces?1306

In this case, the probability of the second card being an ace is affected by what the first card was,1310

because it can change the number of aces in the deck.1317

If we pull out a card that is an ace on the first one, then there is now one fewer ace for our second pull.1319

If we don't pull out an ace on the first one, then there is the same number of aces in our second pull.1326

So, what we do on that first pull affects what will happen in the second pull.1331

We denote conditional probability with this sort of symbols, this notation.1336

The conditional probability of B occurring if A does occur is p of B bar A n things.1343

So, that is how we would say it--just the symbols.1352

If we want to talk about it, it would be the conditional probability of B occurring if A occurs.1354

Or we could also say this as the probability of B occurring, assuming that A occurs.1358

So, if we can know for sure that A will occur, then what is B's chance of occurring, with that piece of information already in mind?1363

This is the idea of conditional probability.1371

We assume the second one; the second one is the assumed one, and the first one is what we are looking at.1374

We assume the second one, p of B, A, the conditional probability of B occurring if A does, so B bar A.1391

A is going to be assumed; the second thing that shows up is assumed.1398

The first thing is now what we are trying to figure out the probability of, if we can have that assumption.1401

All right, so in the above example, we could denote the probability of the second ace being pulled, if the first ace has already been pulled.1405

What is the chance of that second ace coming out, if the first ace has already come out?1412

Then, we have our assumed thing, the first ace; and what we are looking for now is1417

the probability of that second ace, with that assumption there: the probability of the second ace, assuming a first ace.1423

Given two events A and B, where the outcome of A affects the outcome of B,1431

the probability of both events occurring is: probability of A and B is equal to the probability of A occurring on its own,1435

multiplied by the probability of B occurring, assuming that A occurs (this conditional probability).1443

Let's see this as an example: in the previous example, the result of one card1450

affects the result of the second card, because it changes what is in the deck.1453

So, the probability of that very first ace is 4 out of 52, because there are 4 aces in the deck when we pull it out, and there are 52 cards total to pull from.1457

But for the second ace, there is now one less ace, and there is one less card, if we assume that an ace was pulled on that first draw.1468

So, there is one less ace; that means that we now have 3 aces to pull from,1477

divided by 51 cards that we are pulling from total, because now there is just one less card in the deck.1481

Thus, the probability for drawing a pair of aces--if we were looking for the probability of both of these happening,1486

the first ace and a second ace pulled on two cards, then the probability of that first ace,1492

times the probability of the second ace, assuming a first ace has already been pulled, is how we get this.1498

We multiply those two together: so the probability of the first ace was 4/52, and then we multiply it1505

by the probability of the second ace, if we can assume that the first ace has already happened.1511

4/52 times 3/51...that comes out to be 1/221.1516

All right, that idea of conditional events is really interesting stuff.1524

But it is also probably the most extreme stuff that you would end up seeing on a test.1527

I would doubt that you would even see that very often, but it is pretty cool; we will see that in the final example.1532

All right, let's start with a nice, simple example: Given a bag containing 4 red marbles, 7 blue, 11 green, and 2 purple,1536

what is the chance of a blue marble being drawn randomly from the bag?1543

We are looking for what the probability is of a blue coming out.1547

How do we figure out probability? Well, we know that we are drawing randomly.1553

So, if it is randomly, we know that all of the possibilities are equally likely; that is what that "random" word means:1558

that it is a tip that says that all of these are equally likely;1563

you don't have to worry about it, which means that we can use that nice, simple formula.1566

So, it is the number of ways that the event can occur (in this case, the number of blues we could pull out),1569

the number of blue marbles, divided by the number of marbles total, all of the ways that something can happen.1575

In this case, we know...how many marbles are there that are blue? 7 are blue.1587

So, it is 7 divided by...how many marbles do we have total?1592

We have 4 red; we have to include the 7 (they are still 7, so they count as some of them), plus 7, plus 11, plus 2.1595

That is all of the marbles totaled together: we work that out: we have 7/(4 + 7 + 11 + 22), or 24; so 7/24.1605

We can't simplify it any more; and there is our probability.1615

A 7 out of 24 chance exists of pulling a blue marble randomly from the bag.1619

The second example: a class has a breakdown in grades shown by the table below.1624

If a student is selected randomly from the class (once again, there is that word "randomly";1628

it means that we can just assume that everything is equally likely), what is the chance that they have a B or a C?1632

If we are doing this, the probability of a B or a C...well, we can also look at that as the union, B union C.1638

It can be B, or it can be C; so we can add these together.1651

That would be equal to the probability of a B, plus the probability of a C, because it can be either B or C.1654

And we know that they are mutually exclusive; we don't have to worry about them being B and C simultaneously.1662

So now, we can work this out: what is the probability of a B?1667

Well, there are 12 students that have B's; let's expand a little bit more...1669

Here is the probability of B's, and then we will have the probability of C's over here.1675

How many students have B's? We have 12 students with B's, divided by...how many students do we have total?1680

We have 8 here, 12 here, 5 here, 7 here; so we add them all together to figure out how many students we have in the class.1686

8 + 12 + 5 + 7; then we add...what is the probability of a C student?1692

Well, a C student has 5...there are 5 C students total in the class.1697

And then, once again, we divide it by the number of students total, 8 + 12 + 5 + 7.1701

We work this out: 12 over...8 + 12 is 20, plus 5 is 25, plus 7 is 32, plus 5/32, equals 17/32.1707

We can't simplify that any more, so there is our probability.1721

If we draw a student randomly from the class, we have a 17/32 chance of pulling a B or a C student.1724

Next, a bag of marbles contains 12 red marbles, along with various others.1733

If you draw a marble from the bag at random, you have a 15% chance of drawing a red one.1738

What is the total number of marbles in the bag?1744

So, for this, we have 12 red marbles; we know that, if we pull out of the bag at random, we have a 15% chance of drawing a red one.1746

So, our very first thing is that we want to turn 15% as a chance...we can't really work with percentages.1752

We have seen this before; we have to turn them into decimal numbers before we can work with them in math, usually.1758

15% we change into 0.15; so it is a 0.15 probability.1763

Now, if we work this our normal way, then we know that the probability of pulling out a red1769

is equal to the number of reds in the bag, divided by the number of marbles total.1775

The total number is on the bottom.1788

What is the probability of pulling out a red?1790

Well, that is a 0.15; how many red are there? There are 12 red marbles.1792

So, it is 12 divided by the number total; the number total is just some number.1798

If we wanted, we could replace it with x, or whatever we felt like.1805

I am just going to keep writing "number total," because we know that it is just a number that we are working with.1808

We multiply both sides by number total; number total will cancel out on the denominator on the right, and appear on the left.1813

And we divide both sides by 0.15; so now, we have 12 over 0.15.1818

We use a calculator to figure out what is 12 divided by 0.15; that comes out to be 80.1824

So, we know that the total number of marbles in the bag is 80 marbles, because we were guaranteed1831

that if we pull out randomly from the bag, there is going to be a 15% chance of drawing a red one.1836

And we knew how many red marbles we started with.1842

It is a probability, with just a slight spin of algebra on it.1844

The fourth example: If you roll three fair six-sided dice ("fair" means that all of the sides are equally likely),1848

what is the chance of a 6 coming up on each die? on none of the dice? and finally, on at least one of the dice?1855

So first, let's say we are looking at all dice.1862

6 on all: if it is going to be a 6 on all of them, then it is a question of what the first one is.1866

Well, the second one is an independent event of the first one.1875

And the third one is an independent event of the first two.1878

One die's result does not affect the result of the next die, which does not affect the result of the next die.1881

They are independent events; so what is the chance of one die coming up as a 6? It is 1 out of 6.1885

What is the chance of the next die coming up as a 6? It is 1 out of 6.1893

The chance of the next die coming up as a 6 is 1 out of 6.1896

By our rule about independent events, it is multiplying them together.1899

If they are independent events, and you want to know what the probability of them all occurring is, just multiply all of the probabilities together.1903

1/6 times 1/6 times 1/6...we work that out; that comes out to 1/216; so it is a 1 out of 216 chance of a getting a 6 on all of the dice.1909

Next, what if we want to do none of the dice? 1921

If it is none of the dice, what is the chance of getting no 6's on any die?1926

Well, that would be 1, 2, 3, 4, 5...those are the things that are allowed to come up.1933

So, we have 5 possibilities, divided by 6, the total number of things that could happen.1937

And that is going to be the same for the second die and the third die: 5/6 times 5/6 times 5/6.1942

We could also write this as 53, divided by 63.1949

We work that out with a calculator, and we get 125...actually, we probably don't need a calculator for that...divided by 216.1953

So, the chance of getting no 6's on any dice is 125/216.1961

Finally, at least one of the dice: this might be the one that seems hardest at first,1968

but it is actually easy, once we know this part right here, if we know none of them.1974

So, remember what we talked about before with the complement of an event.1979

If an event does not happen, then we are talking about the event complement happening.1982

If E does not happen, then Ec does happen.1988

If none of the dice happens, then there are no 6's on any of them.1992

But if none of the dice does not happen--that is to say, we don't roll a 6 on none of the dice--1996

then that means we have rolled a 6 on one of the dice, or more.2002

So, at least one is going to be the probability of the complement to the none event.2005

So, what we talked about before was the probability of none, complement: it is equal to 1 - the probability of none.2010

Another way of looking at it is just that, if you know what the probability of an event is,2023

then the probability of the opposite thing happening is 1 minus the probability of that event; that is what this none complement thing is.2027

It is the opposite of that event occurring.2033

We know 1 minus...we just figured out what is the probability of none occurring; it is 125 over 216.2036

So, we get that 91 out of 216 is the chance that at least one of the dice will come up with a 6 on it.2043

The final example: A poetry class has 17 boys and 13 girls.2054

If the teacher randomly selects 4 students from the class, what is the chance that they will all be boys?2059

This is the idea of conditional probability.2065

We have some first thing that is going to happen; but then, the second thing is going to also be affected by that first thing.2068

And then, the third thing is going to be affected by that second, and the first, thing.2076

And then, the fourth thing is going to be affected by that first, second, third thing as well.2081

So, what we do is figure out the probability of the first thing; then we multiply it by the probability of the second thing, assuming that first thing.2086

And then, we multiply that by the probability of the third thing, assuming those first two things.2091

And then, we multiply that by the probability of the fourth thing, assuming those first three things.2094

That is how conditional probability worked when we talked about it earlier.2098

So, if we start with how many students there are total in the class...if we have 17 boys and 13 girls, we assume we have 30 total.2101

So, if we have 30 total, then for the first one, we have 17 out of 30.2110

But for the next one, we pull out one of the students.2116

We have pulled out one boy; so that means we now only have 16 boys.2120

How many students do we have total? Our total of students has also gone down by 1, because we have already used one of the students.2124

We pulled out a boy; he is still one of the students in the class, so we now reduce from 30 to 29 students in the class.2130

Next, we pull out another boy; we are going to now be at 15 boys, divided by...we pull out another student...28.2136

Finally, our fourth boy: we are now at 14 boys left to pull from, and we are now at 27 students total in the classroom, after our three pulls so far.2142

If we want to figure this out, conditional probability is that we just multiply them all together: 17/30 times 16/29 times 15/28 times 14/27.2151

We work that out, and that ends up simplifying to the not-that-simple-looking 68/783, which comes out approximately to 0.087.2165

We have a little bit less than a 10% chance, a .087 chance, of managing to pull all boys if we pull 4 students.2178

It drops down pretty quickly; the first boy has a 17/30 chance, but we drop down pretty quickly by the time we are at the fourth boy.2186

It is a less-than-one-in-ten chance.2192

All right, that finishes all of our stuff about combinatorics in here, our ideas of counting, permutations, combinations, and probability.2194

I hope you have a reasonable grounding; this is all of the basics that you need for this level of math.2201

But there is a huge, huge amount of stuff to explore out there.2204

If you thought that this stuff was interesting, just do a quick search on combinatorics.2207

You will find out all sorts of cool things; there are all sorts of really cool things in combinatorics--2210

how to count things; there are lots of cool ideas in that.2214

All right, we will see you at Educator.com for the next lesson--goodbye!2216