[T1.5] Statistical reasoning

Conditional probability

Suppose I pick a card at random from a pack of playing cards, without showing you. I ask you to guess which card it is, and you guess the five of diamonds. What is the probability that you are right? Since there are 52 cards in a pack, and only one five of diamonds, the probability of the card being the five of diamonds is 1/52. Next, I tell you that the card is red, not black. Now what is the probability that you are right? Clearly you now have a better chance of being right than you had before. In fact, your chance of being right is twice as big as it was before, since only half of the 52 cards are red. So the probability of the card being the five of diamonds is now 1/26. What we have just calculated is a conditional probability--the probability that the card is the five of diamonds, given that it is red.

If we let A stand for the card being the five of diamonds, and B stand for the card being red, then the conditional probability that the card is the five of diamonds given that it is red is written $P(A\vert B)$ . The definition of conditional probability is:

$\begin{displaymath}P(A\vert B) = \frac{P(A \mbox{ and } B)}{P(B)} \end{displaymath}$

In our case, $P(A \mbox{ and } B)$ is the probability that the card is the five of diamonds and red, which is 1/52 (exactly the same as

, since there are no black fives of diamonds!).

, the probability that the card is red, is 1/2. So the definition of conditional probability tells us that $P(A\vert B)$ = 1/26, exactly as it should. In this simple case we didn't really need to use a formula to tell us this, but the formula is very useful in more complex cases.

If we rearrange the definition of conditional probability, we obtain the multiplication rule for probabilities:

$\begin{displaymath}P(A \mbox{ and } B) = P(A\vert B)P(B) \end{displaymath}$

One might have expected that the probability of A and B would be obtained by simply multiplying the probabilities of A and B, but in fact this only works in special cases. For example, suppose A stands for "the person speaks Cantonese" and B stands for "the person is from Hong Kong". Suppose we pick a person at random from the world's population, and ask what the value of $P(A \mbox{ and } B)$ is. The probability that the person speaks Cantonese is small--the proportion of Cantonese speakers in the world is about 0.01. The probability that the person is from Hong Kong is even smaller--about 0.001. If we multiplied these probabilities together, we would get 0.00001, but this is clearly the wrong way to calculate the probability that the person is both from Hong Kong and a Cantonese speaker.

If we use the definition of conditional probability, we can see the mistake. $P(A\vert B)$ is the conditional probability that a person speaks Cantonese given that they're from Hong Kong. This number is close to 1. So the correct estimate of the value of $P(A \mbox{ and } B)$ is about the same as , the probability that the person is from Hong Kong.

If instead A stands for "the person is female" and B stands for "the person was born in March" then the situation changes. The probability that a person picked at random is female is roughly 1/2, and the probability that the person was born in March is roughly 1/12. The probability $P(A \mbox{ and } B)$ that the person is both female and born in March is about 1/24, since about half the people born in March are female. In this case, the probability of A and B is obtained by multiplying the probabilities of A and B. The difference between this case and the last one is that a person's sex and birth date are independent (as far as I know!), whereas a person's native language and where they come from are clearly not independent.

In terms of the multiplication rule, if A and B are independent, then the conditional probability $P(A\vert B)$ is the same as . (The probability that a person is female given that they were born in March is just the same as the probability that the person is female.) So for independent events, we have a special multiplication rule:

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

We (implicitly) used the special multiplication rule earlier on, when we calculated that the probability that five tossed coins all show tails is 1/2 $\times$ 1/2 $\times$ 1/2 $\times$ 1/2 $\times$ 1/2 = 1/32. In doing so, we assumed that the results of the five tosses are all independent of each other.

Self-test questions:

Suppose you throw two dice, one after the other. What is the probability that the first die shows a 2?
What is the probability that the second dice shows a 2?
What is the probability that both dice show a 2?
What is the probability that the dice add up to 4?
What is the probability that the dice add up to 4 given that the first die shows a 2?
What is the probability that the dice add up to 4 and the first die shows a 2?