[T1.4] Statistical reasoning

Combining probabilities

Suppose I roll two dice. What is the probability that I will get at least one 6? I might reason as follows: For each die, the probability of rolling a 6 is 1/6. For two dice, the probability of getting at least one 6 is the probability that the first one is a 6 plus the probability that the second one is a 6. That is, the probability of at least one 6 is 1/6 + 1/6, which is 1/3.

But there's clearly something wrong with that reasoning. Think about what would happen if I extended that reasoning to rolling six dice; the reasoning tells me that the probability of getting at least one 6 is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 +1/6, which is 1. But that's not true; it's not certain that I'll get a 6. (And if I roll seven dice, the same reasoning tells me that the probability of getting at least one 6 is greater than 1, which is nonsense!)

So what went wrong? Think about all the possible outcomes when you roll two dice. Since there are six possible outcomes for the first die and six possible outcomes for the second, there are 6 $\times$ 6 = 36 possible outcomes overall. Of those 36, six are outcomes in which the first die shows a 6, and six are outcomes in which the second die shows a six (count them!). But that doesn't mean that there are twelve outcomes overall in which one or more dice shows a 6. Why not?

The problem is that we counted one of the outcomes twice, namely the outcome in which both dice show a 6. So in fact only eleven of the 36 outcomes are ones in which one or more dice shows a 6. So the real probability of rolling at least one 6 is 11/36, not 1/3.

Suppose we use the letter A to stand for the first die showing 6, and the letter B to stand for the second die showing 6. In the above discussion, we have been considering the outcome in which either the first die shows a 6 or the second die shows a 6 (or both). We can write this outcome as "A or B". Then the probability we have been looking at, the probability that at least one of the dice shows 6, can be calculated using the following formula:

$\begin{displaymath}P(A \mbox{ or } B) = P(A) + P(B) - P(A \mbox{ and } B) \end{displaymath}$

This says that the probability of either A or B (or both) occurring is the probability of A plus the probability of B, minus the probability of both A and B occurring (to avoid double counting). This formula can be applied to any two events, and is called the addition rule.

Now let's look at a slightly different case. Suppose you're at the racetrack, and you believe that the horse Anova has a probability of 1/9 of winning the next race and the horse Blaise has a probability of 1/3 of winning. What is the probability that either Anova or Blaise wins? According to the addition rule, it's the probability of Anova winning, plus the probability of Blaise winning, minus the probability of both Anova and Blaise winning. But since it's not possible for two horses to win the same race, the probability of both horses winning is zero. So in this case, we can calculate the probability for either Anova or Blaise winning by simply adding the probabilities for Anova winning and for Blaise winning.

Two events which cannot both occur are called mutually exclusive. For mutually exclusive events, like horses winning a race, we don't have to worry about the double counting problem that we discussed earlier, and we can ignore the last term in the addition rule. This gives us the special addition rule for mutually exclusive events:

$\begin{displaymath}P(A \mbox{ or } B) = P(A) + P(B) \end{displaymath}$

Now suppose I throw five coins in the air. What's the probability that at least one of them will show heads? To calculate this probability, we could use the addition rule over and over again, but this gets rather complicated. A much simpler approach is to calculate the probability of this event not happening--that is, the probability that none of the coins will show heads. Since the probability of a single coin showing tails is 1/2, the probability of all five coins showing tails is 1/2 $\times$ 1/2 $\times$ 1/2 $\times$ 1/2 $\times$ 1/2 = 1/32. The outcomes in which at least one coin shows heads include all the possible outcomes except the one in which I get five tails. So since the probabilities for all the possible outcomes must add up to 1, the probability that at least one coin shows heads is 1 minus the probability of getting five tails (11/32 = 31/32).

In general, the probability of an event not occurring is 1 minus the probability of the event occurring. We can express this as a subtraction rule:

$\begin{displaymath}P(\mbox{not } A) = 1 - P(A) \end{displaymath}$

Self-test questions:

Consider an ordinary pack of 52 playing cards. You draw a card at random from the pack. What is the probability that it is a queen?
What is the probability that it is either a queen or a black ace?
What is the probability that it is either a queen or a heart?
What is the probability that it is neither a queen nor a heart?