Section 4.2 Solutions

4.10

1. S = {female, male}
2. S = {6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
3. S = {all numbers between 2.5 and 6 liters per minute}
4. S = {all whole numbers between 150 and 220}

4.12

1. The sample space is shown below.

S = { (1,1), (1,2), (1,3), (1,4)

(2,1), (2,2), (2,3), (2,4)

(3,1), (3,2), (3,3), (3,4)

(4,1), (4,2), (4,3), (4,4)}

2. S = {3, 4, 5, 6, 7, 8, 9}

4.19

P(Yankees) = 0.6

P(Red Sox) = P(Angels)

P(Athletics) = P(White Sox) = (1/3)P(Red Sox)

Let x = P(Red Sox). Then, because all the probabilities must sum to 1,

0.6 + x + x + (1/3)x + (1/3)x = 1, which means x = 0.15.

Therefore,

P(Yankees) = 0.6

P(Red Sox) = P(Angels) = 0.15

P(Athletics) = P(White Sox) = 0.05

4.20

P(3) = P[(1,1)] = 1/16

P(4) = P[(1,2) or (2,1)] = P[(1,2)] + P[(2,1)] = 1/16 + 1/16 = 2/16

P(5) = P[(1,3) or (2,2) or (3,1)] = P[(1,3)] + P[(2,2)] + P[(3,1)] = 1/16 + 1/16 + 1/16 = 3/16

P(6) = P[(1,4) or (2,3) or (3,2) or (4,1)] = P[(1,4)] + P[(2,3)] + P[(3,2)] + P[(4,1)] = 1/16 + 1/16 + 1/16 +1/16 = 4/16

P(7) = P[(2,4) or (3,3) or (4,2)] = P[(2,4)] + P[(3,3)] + P[(4,2)] = 1/16 + 1/16 + 1/16 = 3/16

P(8) = P[(3,4) or (4,3)] = P[(3,4)] + P[(4,3)] = 1/16 + 1/16 = 2/16

P(9) = P[(4,4)] = 1/16

4.22

A = {person is Hispanic}

B = {person is white}

1. P(S) = 0.000 + 0.003 + 0.060 + 0.062 + 0.036 + 0.121 + 0.691 + 0.027 = 1, so this is a legitimate assignment of probabilities.

2. P(A) = P(person is Hispanic) = P[(Hispanic and Asian) or (Hispanic and black) or (Hispanic and white) or (Hispanic and other)] = 0.000 + 0.003 + 0.060 + 0.062 = 0.125.

3. The complement of B is the event that the randomly chosen person is not white (i.e., is Asian or black or other). Note that P(B) = P(person is white) = P[(white and Hispanic) or (white and not Hispanic)] = 0.06 + 0.691 = 0.751. Then,

4. {Person is a non-Hispanic white} = {non-Hispanic and white} = and B}, and Pand B}= 0.691.

4.29

P(lost site) = 0.13

P(good site) = 1 – 0.13 = 0.87

Assuming the seven sites are independent,

P(1^st good and 2^nd good and 3^rd good and 4^th good and 5^th good and 6^th good and 7^th good) =

P(1^st good)P((2^nd good)P(3^rd good)P(4^th good)P(5^th good)P(6^th good)P(7^th good) =

(0.87) (0.87) (0.87) (0.87) (0.87) (0.87) (0.87) = 0.3773.

4.33

a. Because the events are independent, P[(goes up first year) and (goes up second year) and (goes up third year)] = (.65)(.65)(.65) = 0.27

b. 0.35 (because of independence)

c. P(portfolio moves in the same directions in both of the next two years) = P[(up 1^st year and up 2^nd year) or (down 1^st year and down 2^nd year)] = P(up 1^st year and up 2^nd year) + P(down 1^st year and down 2^nd year), because the event are disjoint. Also, P(up 1^st year and up 2^nd year) + P(down 1^st year and down 2^nd year) = P(up 1^st year)P(up 2^nd year) + P(down 1^st year)P(down 2^nd year), because the events are independent. Therefore, the probability we are interested in is (.65)(.65) + (.35)(.35) = 0.545.