Homework 10 Solutions (Section 12.2)

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1 Homework 0 Solutions (Section 2.2 December, 203 P (A c P (A P (A B P (A + P (B P (A B ( n n! k k!(n k! 0. Exercise Determine the sample space for the experiment. An urn contains six balls numbered -6 and the experiment consists of selecting ve balls simultaneously without replacement. After selecting ve balls, that means one ball isn't selected. Thus the sample space would be the following: Ω {{2, 3,, 5, 6}, {, 3,, 5, 6}, {, 2,, 5, 6}, {, 2, 3, 5, 6}, {, 2, 3,, 6}, {, 2, 3,, 5}}. Any one of these choice of ve balls can occur when selecting ve balls simultaneously without replacement. 0.2 Exercise 8 Ω {, 2, 3,, 5, 6}, A {, 3, 5}, and B {, 2, 3}. Are A and B disjoint? No, the sets A and B have an event in common. Namely, their intersection A B {3}. 0.3 Exercise 0 Ω {, 2, 3,, 5}, P ( 0., P (2 0.2, and P ( P ( Furthermore, A {, 3, 5} and B {2, 3, }. Find P (A and P (B. We deduce from the probabilities given, that P ( Thus P (A P ( + P (3 + P ( and P (B P (2 + P (3 + P ( 0.3.

2 0. Exercise 2 Ω {, 2, 3,, 5}, P ( 0., P (2 0.2, and P ( P ( Furthermore, A {, 3, 5} and B {2, 3, }. Find P (A B. A B {, 3, 5} {2, 3, } {, 2, 3,, 5} Ω. Thus P (A B P (Ω. 0.5 Exercise Ω {, 2, 3, }, P ( 0., A {2, 3}, B {3, }, P (A 0.7, P (B 0.5, C {, 2}. Find P (C. The short solution is to notice that C B c. Thus P (C P (B c P (B Exercise 6 Assume that P (A B c 0., P (B A c 0.5, and P ((A B c 0.2. Find P (A B. Let us rst look at diagrams representing these various sets. (a Depiction of A B c (b Depiction of B A c (c Depiction of (A B c (d Depiction of A B Figure : A is represented by the circle on the left and B is represented by the circle on the right and Ω is the entire rectangle. 2

3 After thinking about the information given to us, we might realize that A B is a disjoint union of the sets A B c, B A c, and A B. Because the union is disjoint, we have P (A B P (A B c + P (B A c + P (A B. Since P (A B 0.2, we have P (A B and conclude that P (A B Exercise 8 Assume that P (A 0., P (B 0., and P (A B 0.7. Find P (A B and P (A c B c. Because we have so that P (A B P (A + P (B P (A B P (A B P (A B 0.. Next, drawing the picture for A c B c brings us to the realization that A c B c (A B c. Figure 2: Depiction of A c B c. A is represented by the circle on the left and B is represented by the circle on the right and Ω is the entire rectangle. Therefore, P (A c B c P (A B

4 0.8 Exercise 22 Find the probability of no heads when tossing three fair coins. The outcomes of tossing three fair coins occurs equally likely and we solve this problem using the material of this section. Letting A {no heads} {all three are tails}, we have P (A A Ω Exercise 2 Find the probability of three or more heads when tossing four fair coins. This outcome occurs when there are exactly three heads or when there are exactly four heads. Thus A {three or more heads} {THHH, HTHH, HHTH, HHHT, HHHH} and we have P (A A Ω Exercise 26 Roll two fair dice and nd the probability that the sum of the two number is even. Updated Solution: The sum is even when both die are even or both are odd. The possibilities to draw two even are ( 3 ( 3 9. Similarly, the number of ways to have the two die be odd is ( ( In total, we have 8 outcomes out of 36 possible outcomes and conclude the probability that the sum is even is Exercise 28 Roll two fair dice and nd the probability that the minimum of the two numbers will be greater than. Here we have A {(,, (, 5, (, 6, (5,, (5, 5, (5, 6, (6,, (6, 5, (6, 6} and obtain P (A A Ω 9 36.

5 0.2 Exercise 3 Three children. Suppose a child has a fty percent chance of being male or female. What is the probability the family has at least one boy? Let A {at least one boy}. Then A c {no boys} {FFF}. Then P (A c Ac Ω 8 and so P (A P (A c Exercise 36 Color blindness is an X-linked inherited disease. A women with the color blindness gene on one of her X chromosomes, but not the other, has normal vision. A man who carries the gene on his only X chromosome is color blind. If a woman with normal vision who carries the gene on one of her X chromosomes has a child with a man who has normal vision, what is the probability that their child has normal vision? The female has X normal X colorblind and the male has X normal Y. Thus their possible ospring are and we have X normal X colorblind X normal X normal X normal X normal X colorblind Y X normal Y X colorblind Y P (child is colorblind 0. Exercise 0 Five blue and three green balls in an urn. Remove three without replacement. What is the probability that at least two of the three balls are green? Let us count the ways we can have exactly two green balls. We have two green and one blue, and have to remove the ordering of the green balls. We have ( 3 2 ( Similarly, the number of ways we can have exactly three green balls is ( Thus the probability of removing at least two green balls is 6 (

6 0.5 Exercise 2 Select ve cards without replacement. What is the probability you get four aces. We can think of this problem as similar to Exercise 0. Here, instead of green balls and red balls, we have cards which are Aces and cards which are not. Then to choose four the number of ways is ( The total number of combinations is ( 52 5 that the probability of getting four aces is 0.6 Exercise ( so 5, 5. Three green, ve blue, and four red. Take three balls without replacement. What is t he probability that all three balls are the same color? This event is a union of three dierent events, having all three green, all three blue, and all three red. The corresponding totals are ( 3 3! 0!3!, ( 5 5! 2!3! 0, and (!!3!, respectively. There are a total of 5 events that we are counting out of ( 2 2! 9!3! 220. Thus the probability of getting all three balls the same color is 0.7 Exercise Four cards. What is the probability that all are of dierent suits? This probable is similar Exercise 0 and Exercise 2, but instead of green balls and red balls or Aces and not aces, we have diamonds, clubs, hearts, and spades. We choose one of each, so we have ( 3 ( 3 ( 3 ( 3 3. The number of four card hands is ( 52. Thus, we have the probability of getting all four of dierent suits is 3 ( , , 075 6