We have now studied all the rules of probability for discrete events: the axioms of probability and conditional probability and the definition of independence. The following are examples of calculating probabilities using these rules.
EXAMPLE 2.5.1 Using the axioms of conditional probability, we shall solve the same problem that appears in Example 2.3.4. In the present approach we recognize only two types of cards, aces and nonaces, without paying any attention to the other characteristics—suits or numbers. We shall first compute the probability that three aces turn up in a particular sequence: for example, suppose the first three cards are aces and the last two nonaces. Let A* denote the event that the ith card is an ace and let TV, denote the event that the ith card is a nonace. Then, by the repeated application of Theorem 2.4.1, we have (2.5.1) P(Ax П A2 П A3 П Л/4 П -(V5)
= P(2V51 Ax П A2 П A3 П N4)P(Ai П A2 П A3 П iV4)
= P(N5 I Ax П A.2 П A3 П N4)P(N41 Ах П A2 П A3)
• P(Ax П A2 П A3)
= P(Aj)P(A2 I Ax)P(A3 I Aj П A2)P(N4 | Aj П A2 П A3)
• P((V51 Aj П A2 П A3 П N4)
= A JL A 48 47 52 ‘ 51 ’ 50 ‘ 49 ‘ 48 ‘
There are C3 ways in which three aces can appear in five cards, and each way has the same probability as (2.5.1). Therefore the answer to the problem is obtained by multiplying (2.5.1) by C3.
EXAMPLE 2.5.2 Suppose that three events, A, B, and C, affect each other in the following way: P(B С) = У2, P(P С) = У3, P(A | В) = У2, P(A | B) = У3. Furthermore, assume that P(A | В П С) = P(A В) and that P(A I В П С) = P(A I B). (In other words, if В or В is known, C or C does not affect the probability of A.) Calculate P(A | C) and P(A | C).
Since A = (А П B) U (А П 1), we have by axiom (3) of conditional probability
(2.5.2) P(A I С) = P(A П В I C) + P(A П В C).
By repeated application of Theorem 2.4.1, we have
(2.5.3) P(A П В I С) = П B П C^
_ P(A I P П C)P(B П C)
= P(A I P П C)P(P I C).
Therefore, by our assumptions,
(2.5.4) P(AHBQ = P(A P)P(P.| C) = | | ^ ■
(2.5.5) P(A П В I C) = P(A I B)P(B I Q = | = І’
3 2 6
Finally, from (2.5.2), (2.5.4), and (2.5.5) we obtain P(A C) = 5/12. Calculating P(A I C) is left as a simple exercise.
EXAMPLE 2.5.В Probability calculation may sometimes be counterintuitive. Suppose that there are four balls in a box with the numbers 1 through 4 written on them, and we pick two balls at random. What is the probability that 1 and 2 are drawn given that 1 or 2 is drawn? What is the probability that 1 and 2 are drawn given that 1 is drawn?
By Theorem 2.4.1 we have
P( 1 and 2) P(1 or 2)
P(1 and 2)
= 1 ~ 3 ’
The result of Example 2.5.3 is somewhat counterintuitive: once we have learned that 1 or 2 has been drawn, learning further that 1 has been drawn does not seem to contain any more relevant information about the event that 1 and 2 are drawn. But it does. Figure 2.1 illustrates the relationship among the relevant events in this example.
EXAMPLE 2.5.4 There is an experiment for which there are three outcomes, A, B, and C, with respective probabilities pA, pB, and pc – If we try
3 and 4
figure 2.1 Characterization of events
this experiment repeatedly, what is the probability that A occurs before В does? Assume pc Ф 0.
We shall solve this problem by two methods.
(1) Let Ai be the event that A happens in the zth trial, and define Bt
and Ci similarly. Let P be the desired probability. Then we have (2.5.8) P = P(Ai) + P(Ci П Ai) + P(Ci П C2 П A3) + . . .
= P(Aj) + P(C)P(Ai) + P(Ci)P(Cz)P(A3) + …
= pA + pCpA + pCp A + • • •
(2) We claim that the desired probability is essentially the same thing as the conditional probability of A given that A or В has occurred. Thus
which gives the same answer as the first method. The second method is an intuitive approach, which in this case has turned out to be correct, substantiated by the result of the rigorous first approach.
example 2.5.5 This is an application of Bayes’ theorem. Suppose that a cancer diagnostic test is 95% accurate both on those who do have cancer and on those who do not have cancer. Assuming that 0.005 of the popu
lation actually has cancer, compute the probability that a particular individual has cancer, given that the test says he has cancer.
Let C indicate the event that this individual actually has cancer and let T be the event that the test shows he has cancer. Then we have by Theorem
0. 95 • 0.005
0.95 • 0.005 + 0.05 • 0.995 475
1. (Section 2.2)
(a) А П (B U С) = (А П B) U (А П C).
(b) A U (В П C) = (A U В) П (A U C).
(c) (A — С) П (В — С) = (Л П В) — C.
2. (Section 2.3.1)
Complete Example 2.3.2.
3. (Section 2.4.1)
Show that Theorem 2.4.1 implies (2), (3), and (4) of the axioms of conditional probability.
4. (Section 2.4.2)
Suppose that the Stanford faculty consists of 40 percent Democrats and 60 percent Republicans. If 10 percent of the Democrats and 70 percent of the Republicans vote for Bush, what is the probability that a Stanford faculty member who voted for Bush is a Republican?
5. (Section 2.4.3)
Fill each of the seven disjoint regions described in the figure below
by an integer representing the number of simple events with equal probabilities in such a way that
(a) (2.4.8) is satisfied but (2.4.5), (2.4.6), and (2.4.7) are not.
(b) (2.4.5), (2.4.6), and (2.4.7) are satisfied but (2.4.8) is not.
(c) (2.4.5), (2.4.6), (2.4.7), and (2.4.8) are all satisfied.
6. (Section 2.5)
Calculate P(A | C) in Example 2.5.2.
7. (Section 2.5)
If the probability of winning a point is p, what is the probability of winning a tennis game under the “no-ad” scoring? (The first player who wins four points wins the game.)
8. (Section 2.5)
Compute the probability of obtaining four of a kind when five dice are rolled.
9. (Section 2.5)
If the probability of being admitted into a graduate school is 1 /n and you apply to n schools, what is the probability that you will be admitted into at least one school? Find the limit of this probability as n goes to infinity.
10. (Section 2.5)
A die is rolled successively until the ace turns up. How many rolls are necessary before the probability is at least 0.5 that the ace will turn up at least once?