7  Conditional Probability

7.1 Conditional Probability by Counting

Example: A health insurance pool includes 200 individuals. The insurer is interested in the number of smokers in the pool among both males and females. The following table (called a contingency table) shows the desired numbers.

	Males \((M)\)	Females \((F)\)	Total
Smokers \((S)\)	28	22	50
Non-smokers \((S^c)\)	72	78	150
Total	100	100	200

Suppose you were told that the selected individual was a male, and asked for the probability that the individual was a smoker, given that the individual was a male (denoted \(P(S|M)\)). Since there are only 100 males and 28 of them are smokers, the desired probability can be found by dividing the number of male smokers by the total number of males: \[P(S|M) = \frac{n(S\cap M)}{n(M)} = \frac{28}{100} = .28\] This problem can also be solved using probabilities. We can see that: \[P(S|M) = \frac{P(S\cap M)}{P(M)} = \frac{28/200}{100/200} = \frac{.14}{.50} = .28\] ## Defining Conditional Probability For any two events \(A\) and \(B\), the conditional probability of \(A\) given \(B\) is defined as: \[P(A|B) = \frac{P(A\cap B)}{P(B)}\] Example: A card is drawn at random from a standard 52-card deck. The card is not replaced. Then a second card is drawn at random from the remaining cards. Given that the first card drawn was a king (K1), find the probability that the second card is also a king (K2). Solution: the deck will contain 51 cards and only 3 kings for the second draw. Then \[P(K2 | K1) = \frac{3}{51} \approx .0588\]

Proposition (Multiplication Rule for Probabilities):
The definition of conditional probability can be rewritten as a multiplication rule for probabilities: \[P(A\cap B) = P(A|B)P(B)\]

Example: Two cards are drawn from a standard deck without replacement. Find the probability that both are kings.

Solution: \[P(K1 \cap K2) = P(K1) P(K2|K1) = \frac{4}{52} \frac{3}{51} \approx .0045\]