Week Two Assignment Bus308: Statistics for Managers
Week Two Assignment
BUS308: Statistics for Managers
Tiffany Aldridge
January 9, 2012
Week Two Assignment
Chapter 4: 4.4, 4.20
4.4
Suppose that a couple will have three children. Letting B denote a boy and G denote a girl:
a. Draw a tree diagram depicting the sample space outcomes for this experiment
b. List the sample space outcomes that correspond to each of the following events: 1) All three children will have the same gender. BBB, GGG 2) Exactly two of the three children will be girls. BGG, GBG, GGB 3) Exactly one of the three children will be a girl. BBG, BGB, GBB 4) None of the three children will be a girl. BBB c. Assuming that all sample space outcomes are equally likely, find the probability of each of the events given in part b.
1) Probability of all three children having the same gender:
P(BBB)= 1/8
P(GGG)= 1/8
P(BBB) + P(GGG)= 1/8 + 1/8 = 2/8 = 1/4 = .25
2) Probability of exactly two of the three children will be girls:
P(BGG) + P(GBG) + P(GGB) = 1/8 +1/8 + 1/8 = 3/8 = .375
3) Probability of exactly one of the three children will be a girl:
P(BBG) + P(BGB) + P(GBB) = 1/8 + 1/8 + 1/8 = 3/8 = .375 4) Probability that none of the three children will be a girl:
P(BBB) = 1/8 = .125
4.20
John and Jane are married. The probability that John watches a certain television show is .4. The probability that Jane watches the show is .5. The probability that John watches the show, given that Jane does is .7.
a. Find the probability that both John and Jane watch the show.
b. Find the probability that Jane watches the show, given that John does.
c. Do John and Jane watch the show independently of each other? Justify your answer.
No, John and Jane do not watch the show independently of each other. If John watched independently, then= this is not the situation as .7 ≠ .4. If Jane watched independently, then = this is not the situation as .875 ≠ .5.
BUS308: Statistics for Managers
Tiffany Aldridge
January 9, 2012
Week Two Assignment
Chapter 4: 4.4, 4.20
4.4
Suppose that a couple will have three children. Letting B denote a boy and G denote a girl:
a. Draw a tree diagram depicting the sample space outcomes for this experiment
b. List the sample space outcomes that correspond to each of the following events: 1) All three children will have the same gender. BBB, GGG 2) Exactly two of the three children will be girls. BGG, GBG, GGB 3) Exactly one of the three children will be a girl. BBG, BGB, GBB 4) None of the three children will be a girl. BBB c. Assuming that all sample space outcomes are equally likely, find the probability of each of the events given in part b.
1) Probability of all three children having the same gender:
P(BBB)= 1/8
P(GGG)= 1/8
P(BBB) + P(GGG)= 1/8 + 1/8 = 2/8 = 1/4 = .25
2) Probability of exactly two of the three children will be girls:
P(BGG) + P(GBG) + P(GGB) = 1/8 +1/8 + 1/8 = 3/8 = .375
3) Probability of exactly one of the three children will be a girl:
P(BBG) + P(BGB) + P(GBB) = 1/8 + 1/8 + 1/8 = 3/8 = .375 4) Probability that none of the three children will be a girl:
P(BBB) = 1/8 = .125
4.20
John and Jane are married. The probability that John watches a certain television show is .4. The probability that Jane watches the show is .5. The probability that John watches the show, given that Jane does is .7.
a. Find the probability that both John and Jane watch the show.
b. Find the probability that Jane watches the show, given that John does.
c. Do John and Jane watch the show independently of each other? Justify your answer.
No, John and Jane do not watch the show independently of each other. If John watched independently, then= this is not the situation as .7 ≠ .4. If Jane watched independently, then = this is not the situation as .875 ≠ .5.