Stastics

4.4. Suppose that a couple will have 3 children. B = boy & G = 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 3 children will have the same gender. 2) Exactly 2 of the 3 children will be girls. 3) Exactly 1 of the 3 children will be a girl. 4) None of the 3 children will be a girl. C) Assuming that all sample space outcomes are equally likely, find the probability of each of the events given in part b. a) b) 1. BBB, GGG 2. BGG, GGB, GBG 3. BBG, BGB, GBB 4. BBB c) 1. 2/8 = ¼ 2. 3/8 3. 3/8 4. 1/8 4.20 John & Jane are married. The probability that John watches a certain TV 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 & Jane watch the show. b) Find the probability that Jane watches the show, given that John does. c) Do John & Jane watch the show independently of each other' a) P(John│Jane)=P(John ⋂ Jane)/P(Jane) .7=P(John ⋂ Jane)/(.5) P(John ⋂ Jane)=.35 b) P(Jane│John)=P(John ⋂ Jane)/P(John) P(Jane│John)=(.35)/(.4)=.875 c) No, they do not watch the show independently = P(John ⋂ Jane)≠P(John).P(Jane) 5.12 a. p (x) = x/15 The probability distribution of x x 1 2 3 4 5 P (x) 1/15 2/15 3/15 4/15 5/15 b. Properties of a discrete probability distribution pj (x) ≥ 0 ∑ p (x) = 1 Since each individual of probability is greater than zero P (1), P (2), P (3), P (4), P (5) ∑ p (x) = P (1)+ P (2)+ P (3)+ P (4)+ P (5) 1/15 + 2/15 + 3/15+ 4/15+5/15 =15/15=1 c. M = Mean = ((1+2+3+4+5))/5=15/5=3 d. Find the variance of (σ2x) and the standard deviation (σx) ∑ (x2) = (1)2(1/15)+ (2)2(2/15)+ (3)2(3/15)+ (4)2(4/15)+ (5)2(5/15)= (1/15)+ (8/15)+ (27/15)+ (64/15)+ (125/15)= 225/15 = 15 Variance of x = σ2 =∑ (x2) – M2 = 15 – (3)2 = 15 – 9 = 6 Standard deviation of x = σ = √σ 2 = √6 = 2.45 6.22(a)