In a previous post, we discussed this formula:
Probabilities of Compound Events
Let A and B be independent of one another. Then, P(A and B) = P(A)P(B)
An example will help to illustrate why this formula makes sense.
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.
But to help explain why this rule makes sense, we give another way to think about this problem. So there are two possible ways my coin flip can go:
And then, no matter which way my coin faces, there are another six ways the dice can land:
Now, intuitively, each of these outcomes are equally likely. For example, the probability of getting tails and rolling a 4 should be equal to the probability of getting heads and rolling a 2. Then, if we look at all the different outcomes, we find that there are 12 such outcomes and, in exactly one of them, we flip a heads and roll a six.
Thus, following our earlier rule, we get that P(A and B) = 1/12.
Note that it will not matter whether you start out by listing the outcomes of the dice roll or the outcomes of the coin flip:
In either case, we get 1/12.
In general, listing out all the different possible outcomes will be infeasible. But it helps to get a sense for why this formula works. Here is an example where we wouldn’t bother to list out all the different outcomes:
Suppose I roll three fair six-sided die in a row. What is the probability that the first die lands on an even number, and the second and third die both land on odd numbers?
We would not bother to list out all the possibilities, since there would be of them! (Since there are 6 possibilities for the first die, 6 for the second die, and 6 for the third). But, using our rule, since we know that the die rolls are independent of each other, we just need to calculate:
And this is just