1.6 Conditional Probability The probability of an event is often affected by the occurrences of other events. More precisely, probability can change with additional information, and we call these conditional probabilities. Definition: Given two events E and F in a sample space S, we call the probability that E will occur given that F has occurred the conditional probability, and write P (E F ). Read this as the probability of E given F. Example: A card is chosen randomly from a standard deck of 52 cards. (a) What is the probability that this card is an ace? Ans: The sample space S has 52 cards, the event E that the card is an ace has 4 cards so P (E) = n(e) n(s) = 4 52 = 1 13. (b) What is the probability that this card is an ace given that the card is known to be a red non-face card? Ans: Let F be the event that the card is a red non-face card. There are 20 cards in F, and 2 of those are aces, so the required probability is 2 20 = 1 10. The point is that since we already know that the card is a red nonface card, we no longer have to consider the entire sample space. Or in other words, the 20 red non-face cards becomes the new (reduced) sample space. Remark: Observe that the sample space in (b) is smaller than the sample space in (a), resulting in a higher probability in (b). Conditional Probability: Let E and F be two events in a sample space S. Then the conditional probability that E occurs given that F has occurred is defined to be P (E F ) = P (E F ) P (F ) Remark: The events E, F, and E F are all in the original sample space S and P (E), P (F ) and P (E F ) are all probabilities defined on S. However, we can think of P (E F ) as a probability defined on the new sample space F. 1
Example: A new family has moved in next door and is known to have two children. Assume that a boy is as likely as a girl. (a) What the probability that both children are boys? Ans: The sample space is S = {(B, B), (B, G), (G, B), (G, G)} so the required probability is 1 4. (b) What the probability that both children are boys given that at least one is a boy? Ans: Let E be the event that both children are boys, and let F be the event that at least one child is a boy. Then E = {(B, B)}, F = {(B, B), (B, G), (G, B)}, and E F = {(B, B)} so P (E F ) = P (E F ) P (F ) = 1/4 3/4 = 1 3. (c) What the probability that both children are boys given that the elder one is a boy? Ans: Let E be the event that both children are boys, and let G be the event that the elder child is a boy. Then G = {(B, B), (B, G)} and E G = {(B, B)} so P (E G) = P (E G) P (G) = 1/4 2/4 = 1 2. In the formula for conditional probability, if we swap the two events E and F, then the numerator remains the same but the denominator changes, and we get P (F E) = P (E F ) P (E) After rearranging the formulas for P (E F ) and P (F E), we obtain the product rule. Product Rule: If E and F are two events in a sample space S with P (E) > 0 and P (F ) > 0, then P (E F ) = P (F )P (E F ) = P (E)P (F E) 2
Example: Two bins contain transistors. The first bin has 5 defective and 15 non-defective transistors while the second bin has 3 defective and 17 nondefective transistors. If the probability of picking either bin is the same, what is the probability of picking the first bin and a good transistor? Ans: The sample space is S = {1D, 1N, 2D, 2N}. Let E be the event pick the first bin and let F be the event pick a non-defective transistor. Then E = {1D, 1N} and F = {1N, 2N}. The product rule says P (E F ) = P (E)P (F E) = 2 4 P (F E) = 1 2P (F E). P (F E) is the probability of picking a non-defective transistor given that the first bin has been picked so P (F E) = 15 20 = 3 4. Hence P (E F ) = 1 2 3 4 = 3 8. Probability Trees: We can write probabilities on the branches of tree diagrams to illustrate conditional probability. Example: A jar contains 4 red marbles and 6 green marbles. Two marbles are selected randomly from the jar, one at a time and without replacement, and the color of each marble is observed. Construct a probability tree with the information provided: For the second half of the diagram, note that the first marble is not put back into the jar so there are 9 marbles left. That s why the denominators in the second half are all 9. 3
(a) What is the probability that the first marble picked is green and the second marble is red? And translates to intersection. To use the tree diagram to compute probabilities of intersections, multiply along the path through the two events you re intersecting. Ans: 6 10 4 9 = 4 15 (b) What is the probability that the first marble picked is red and the second marble is also red? Ans: 4 10 3 9 = 2 15 (c) What is the probability that the second marble is red? This event is in the second half of the tree diagram. In such cases, consider all paths leading to the desired outcome, multiply along each individual path, then add the results. Ans: 4 3 10 9 + 6 4 10 9 = 2 5 (d) What is the probability that the first marble is green given that the second marble is red? Ans: P (1G 2R) = P (1G 2R) P (2R) = 4/15 6/15 = 4 6 = 2 3 The numerator and denominator were computed in parts (a) and (c) respectively. Independent Events: Two events E and F are called independent events if the outcome of one event does not affect the outcome of the other. In mathematical terms, two events E and F are said to be independent if P (E F ) = P (E) and P (F E) = P (F ) The first formula says that the probability of E occurring is not affected by whether F occurs; the second formula says that the formula of F occurring is not affected by whether E occurs. Combining with the product rule, we have the following theorem, which is a more convenient test for independence of events. Independent Events Theorem: Let E and F be two events with P (E) > 0 and P (F ) > 0. Then E and F are independent if, and only if, P (E F ) = P (E)P (F ) 4
Note: Two events being independent is different from two events being mutually exclusive. Example: Suppose P (A) = 0.3 and P (B) = 0.4. If A and B are independent, find P (A B) and P (A c B c ). Ans: P (A B) = P (A) + P (B) P (A B) = 0.3 + 0.4 P (A B) = 0.7 P (A B). Since A and B are independent, P (A B) = P (A)P (B) = (0.3)(0.4) = 0.12. So P (A B) = 0.7 0.12 = 0.58. By De Morgan s law, A c B c = (A B) c so P (A c B c ) = P ((A B) c ) = 1 P (A B) = 1 0.12 = 0.88. Alternatively, P (A c B c ) = P (A c )+P (B c ) P (A c B c ) = (1 P (A))+(1 P (B)) P (A c B c ) = (1 0.3) + (1 0.4) P (A c B c ) = 1.3 P (A c B c ). Since A and B are independent, A c and B c are also independent, so P (A c B c ) = P (A c )P (B c ) = (1 0.3)(1 0.4) = 0.42. Hence P (A c B c ) = 1.3 0.42 = 0.88. 5
Example: In a medical trial, a new drug was effective for 60% of the patients, and 30% of the patients suffered from a side effect. If 28% of the patients did not find the drug effective nor had a side effect, are the events E (effective) and F (side effect) independent? Ans: P (E) = 0.6, P (F ) = 0.3, and P (E c F c ) = 0.28. P (E F ) = P (E) + P (F ) P (E F ) = 0.9 P (E F ). P (E F ) = 1 P ((E F ) c ) = 1 P (E c F c ) = 1 0.28 = 0.72. Hence P (E F ) = 0.9 0.72 = 0.18. P (E)P (F ) = (0.6)(0.3) = 0.18 = P (E F ) so E and F are independent. The part after this was skipped during lecture. Independent Set of Events: A set of events {E 1, E 2,..., E n } is said to be independent if, for any k of these events, the probability of the intersection of these k event is the product of the probabilities of each of the k events. This must hold for any k = 2, 3,..., n. For example, for the set of events {E, F, G} to be independent, all of the following must be true: P (E F ) = P (E)P (F ) P (E G) = P (E)P (G) P (F G) = P (F )P (G) P (E F G) = P (E)P (F )P (G) Note: If E and F are independent, then so also are E and F c, E c and F, E c and F c. Example: A building has three elevators A, B, and C. The chance that A is not working is 15%, the chance that B is not working is 12%, and the chance that C is not working is 19%. If these probabilities are independent, what is the probability that exactly one elevator is not working? Ans: The probability that only A does not work is (0.15)(1 0.12)(1 0.19). The probability that only B does not work is (0.88)(1 0.15)(1 0.19). The probability that only C does not work is (0.19)(1 0.15)(1 0.12). The probability that exactly one elevator does not work is the sum of these three probabilities, which is 0.28026. Section 1.6 suggested homework: 1, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 41, 43, 45, 47, 49 6