In our previous posts, we talked about the notion of probability, some of its basic features, and how to find the probability of a single event. Here, we will find the probability of a compound event, namely an event where multiple events occur.
For example, we now know that the probability of a fair die landing on 6 is ⅙, while the probability of a fair coin landing heads is ½. But what is the probability that, if I flip a fair coin and roll a fair die, I get that the coin lands heads and the die lands on 6? What is probability that the coin lands heads or the die lands on 6?
To answer that question, we need to introduce a new idea: independence. In ordinary language, talk of “independence” suggests a rebellious child or ideas of liberty and freedom. But here, we use a different notion of independence: two events are independent if neither event affects the likelihood of the other.
An example will help to illustrate the concept. You roll a fair die and then flip a fair coin. We know that, ordinarily, a fair coin has a ½ chance of coming up heads. But now, suppose you knew that the die came up 6. Now, what is the probability that the coin came up heads? Clearly, the answer should remain: ½. The fact that the die came up 6 has nothing to do with the coin! We say that the two events are independent of one another. More precisely, we say:
Definition: Two events, A and B, are independent of one another if:
(i) A occurring does not affect the likelihood of B occurring, and
(ii) B occurring does not affect the likelihood of A occurring
For the purposes of the GRE, it will generally be clear when two events are independent. Here are some standard examples of independence:
- You flip a coin and roll a die. Whether you get heads on the coin ( = Event A) and whether you get 6 on the die ( = Event B) are independent.
- You draw a marble from a bag, replace that marble, and then draw a second marble. Whether the first marble is green ( = Event A) is independent of whether the second marble is purple ( = Event B). Similarly, whether the first marble is green is independent of whether the second marble is green.
Now consider the following case:
A bag contains 10 purple marbles and 7 green marbles. You will randomly draw one marble from the bag and then, without returning the first marble to the bag, you draw a second marble from the bag. Is the event of getting a green marble first independent of the event of getting a purple marble second?
No. This is because whether you get a green marble first or not can actually affect your chances of getting a purple marble second since you do not return the first marble to the bag. Suppose you draw a green marble first. Then, the chance you draw a purple marble second is: (number of purple marbles remaining)/(total number of marbles in bag) = 10/16. But if you did not draw a green marble first, then you must have drawn a purple marble. So then, the odds that you draw a purple marble second is: 9/16.
Here is a more intuitive way to put the same point: suppose your first marble is not green. Then it must be purple (since the bag just has purple and green marbles). That’s one fewer purple marble for you to draw next turn!
The reason why we care about independence is because independent events allow us to easily calculate the probabilities of compound events. What is a compound event?
Definition: Let A and B represent two events. A compound event E is the event where both A and B occur.
For example, suppose I am wondering about the weather this afternoon. Let A = “It rains.” and B = “The Broncos win their game today.” Then, the compound event A and B will be the event where “It rains and the Broncos win.” It will generally be obvious when two events are independent, and sometimes the question will explicitly state that fact. Other examples of independent events include:
- There is a bag with 10 green marbles and 14 yellow marbles. I pick a marble, look at its color, return it to the bag and pick another marble. The color of the first marble is independent of the color of the second.
- I roll a fair die, record its outcome, and then roll it again. The outcome of the first roll is independent of the outcome of the second roll.
- I randomly draw a person's name for a raffle. Then, I flip a coin. The name I draw is independent of my coin flip.
When events are independent, we can calculate the probability of both events occurring via the following rule:
Probabilities of Compound Events
Let A and B be independent of one another. Then, P(A and B) = P(A)P(B)
Let's see this rule in action:
Suppose I roll a fair six-sided die and flip a fair coin. What is the probability that the coin lands heads and the die lands on six?
Following our above rule, we say A = coin lands heads, and B = die lands on six. Then, we can calculate P(A and B) = P(A)P(B). We already know that P(A) = ½ and P(B) = ⅙. So, P(A and B) = 1/12.
Now, this rule may seem a little odd. Why does this rule work? If you're the kind of person who needs to get a sense of why something works in order to learn it, here's an illustration that helps make the rule more intuitive.
Now, here's an application of our rule that uses larger numbers:
You are wondering how poorly your day could go. You know that, at work, one employee (out of 400) will be randomly selected for additional performance reviews. And you know that there is a ⅓ chance that it rains furiously during your commute home. (Of course, whether you are selected or not for the performance review will not affect the weather.) What is the probability that you are picked for the additional performance reviews AND it rains furiously during your commute?
Let A = you are selected for additional performance reviews. Then, P(A) = 1/400 (since one person is randomly selected from a group of 400). Let B = it rains furiously on your way home. The question gives us that P(B) = ⅓. By our rule, P(A and B) = P(A)P(B) = 1/1200. So it is pretty unlikely that your day will be the worst possible.
In our next post, instead of looking at the probability of A and B occurring, we will look at the probability of A or B occurring.
You roll a fair six-sided die. Is the event of you rolling a multiple of 5 independent of you rolling an even number?
No. Suppose you roll a multiple of 5. Then, you must have rolled 5, and 5 is not an even number! So if one event occurs, the other event is impossible. Now suppose you do not roll a multiple of 5. Then, you must have rolled a 1, 2 ,3, 4, or 6. So it is still possible that you rolled an even number, and thus the probability of the second event remains greater than 0. Thus, the first event occurring or not affects the probability of the second event. So they are not independent.
You roll a fair six-sided die. Is the event of you rolling a prime number independent of you rolling an even number?
No. Here, the violation is a little trickier. Suppose you roll a prime number. Then, you must have rolled a 2, 3, or 5. Then, the probability of you rolling an even number is 1/3 since, of those three numbers, only 2 is even. Now suppose you did not roll a prime number. Then, you must have rolled a 1, 4, or 6. Then, the probability of you rolling an even number is 2/3 since, of those three numbers, 4 and 6 are even. So whether the first event occurs or not can affect the probability of the second event. So they are not independent.
There are two events, A and B, which are independent of each other. P(A) = .2 and P(B) = .5. What is P(A and B)?
We can calculate this via our rule: P(A and B) = P(A)P(B) = .2 * .5 = .1.
You flip a fair coin 7 times in a row. What is the probability that all 7 flips come up heads?
Let H1 = the first coin comes up heads, H2 = the second coin comes up heads, ..., H7 = the seventh coin comes up heads. Now, at first blush, it may seem like our rule does not cover this case. But, note that we can define a new event: D = the first two coins come up heads. And D is independent of H3. (Whether the first two coins come up heads or not just doesn't affect whether the third will). So, P(H1 and H2 and H3) = P(D and H3). And we know P(D) = P(H1)P(H2) = 1/2 * 1/2 = 1/4, since the first two coins coming up heads are independent events. So P(D and H3) = P(D)P(H3) = 1/4 * 1/2 = 1/8, since they too are independent. Following this line of reasoning, we get that P(H1 and H2 and ... and H6 and H7) = (1/2)^7 = 1/128.
There are two independent events, A and B. Neither A nor B is guaranteed to happen. P(A) = .7. There are two values:
X = P(A and B)
Y = .7
Which of the following is true?
A. X is greater than Y
B. Y is greater than X
C. X and Y are equal
D. There is not enough information to tell.
B is the correct answer. Since neither event is certain, we know that P(A) < 1 and P(B) < 1. Since A and B are independent, we know that P(A and B) = P(A)P(B) = .7*P(B), since we know the value of P(A). Because P(B) < 1, we know that .7*P(B) < .7. So P(A and B) = .7*P(B) < .7. Thus, X is less than Y.