Math 340 Hints and Solns to Homework Problems ___________________________________________________________________________ Below for the complement of an event A I will use A^c, and for the intersection of two events A and B I will either use AB or (A int B). ____________________________________________________________________________ (p. 45, #11) (a) Let, I = the event that the twins are identical = {B_1 B_2, G_1 G_2} , since they have same sex. F = the event that the twins are fraternal = {B_1 B_2, B_1 G_2, G_1 B_2, G_1 G_2}, since they can different sex in this case. We are given that: a. P(I) = p, b. P(F) = 1-p, c. P(B_1 B_2 | I) = P(G_1 G_2 | I) = 1/2 d. P(B_1 B_2 | F)= P(B_1 G_2 | F)= P(G_1 B_2 | F)= P(G_1 G_2 | F)= 1/4 e. P(B_1 F) = P(G_1 F) = 1/2 f. P(B_2 F) = P(G_2 F) = 1/2 Then, P(B_1 B_2) = P(B_1 B_2 I) + P(B_1 B_2 F) = P(B_1 B_2 | I) P(I) + P(B_1 B_2 | F) P(F) = (1/2)p + (1/4)(1-p) = (1/4)(1+p) (b) P(B_1 G_2) = P(B_1 G_2 | I) P(I) + P(B_1 G_2 | F) P(F) = 0 + (1/4)(1-p) (c) P(G_2 | B_1) = P(G_2 | B_1 F) + P(G_2 | B_1 I) = P(G_2 | B_1 F) + 0 = P(G_2 B_1 F) / P(B_1 F) = P(G_2 B_1 | F) P(F) / P(B_1 F) = (1/4) (1-p) / (1/2) = (1/2) (1-P) (d) P(G_2 | G_1) = P(G_2 | G_1 F) + P(G_2 | G_1 I) = P(G_2 G_1 F)/P(G_1 F) + P(G_2 G_1 I)/ P(G_1 I) =(P(G_2 G_1 | F)P(F)/P(G_1 F)+P(G_2 G_1| I)P(I)/P(G_1 I) = (1/4)(1-p) / (1/2) + (1/2)(p) / (1/2) = (1/2) (1-p) + (1) p = (1/2)(1+p) ____________________________________________________________________________ (p.53, #2) (a) You need to make a tree diagram here. First stage consists of two branches, one for a white draw with prob. 4/10 and one for a black draw with prob. 6/10. The second stage will then consists of two branches for each branch of the first stage. Hence, there should be four branches all together at the end. To calculate the probabilities consider the branch "white and white." This means that we drew a white ball on the first draw, put it back into the urn, and then put 3 more white ball balls into the urn. So, now there are 7 white and a total of 7+6=13 balls in the urn. Therefore, P(white and white) = (4/10)(7/13) P(white and black) = (4/10)(6/13) Likewise you can now get the probabilities for the bottom portion of the tree diagram. Once you have the tree, the rest should be straight forward. ____________________________________________________________________________ (p.54, #7) (a) Let C = the event that you guess correctly. Then, C = BB_1 U WB_3 ,namely, (black and box 1) OR (white and box 3). Then, P(C) = P(B | B_1) P(B_1) + P(W | B_3) P(B_3) = 5/12. Other parts work likewise. ____________________________________________________________________________ (p.70, #3) (a) Let, E = the event that there was at least one head in 3 tosses F = the event that there were at least two tosses Then, P(F | E) = P(FE) / P(E) = P(F)/P(E), since F is a subset of E. Now, P(F) = P(exactly two heads) + P(exactly three heads) = 3 (2/3)^2 (1/3) + (2/3)^3, since P(H)=2/3, P(T)=1/3, and the event "exactly two heads" can occur in 3 ways = (12/27) + (8/27) = 20/27 P(E) = P(exactly one head) + P(exactly two heads) + P(exactly 3 heads) = 3 (2/3) (1/3)^2 + 3 (2/3)^2 (1/3) + (2/3)^3 = (6/27) + (12/27) + (8/27) = 26/27 Hence, the ratio of these two numbers will be the answer: P(F|E) = (20/27) / (26/27) = 20/26 = 10/13. (b) Let, C = the event that there was exactly one head in 3 tosses. Then, P(C | E) = P(CE)/P(E) = P(C)/P(E), since C is a subset of E Now, P(C) = P(exactly one head) = 3 (2/3) (1/3)^2 = 6/27, and the prob of E was computed in (a). Therefore, P(C|E) = (6/27) / (26/27) = 6/26 = 3/13 = 0.2308 ______________________________________________________________________________ (p.71, #4) (a) Let B_i be the event that we get a bell on the i-th wheel, i=1,2,3. Then, P(B_1 B_2 B_3) = P(B_1) P(B_2) P(B_3), by independence = (1/20) (9/20) (1/20) (b) Let, E = the event that we get two bells but not the jackpot. Note that E can occur in three distinct ("disjoint") ways: (i) bells on 1st and 2nd wheel, and a non-bell on third; (ii) bells on 1st and 3rd wheels, and a non-bell on 2nd; (iii) non-bell on 1st and bells on 2nd and 3rd. Using disjointness we find the probability of each case and add them up. Then, P(E) = P(B_1 B_2 B_3^c) + P(B_1 B_2^c B_3) + P(B_1^c B_2 B_3) = 2 (1/20) (9/20) (19/20) + (1/20)^2 (11/20) (c) P(B_1 B_2 B_3) = (3/20)(1/20)(3/20), which is the same as in (a). P(E) = 2 (3/20) (1/20) (17/20) + (3/20)^2 (19/20), which is SMALLER than in (a). Hence, we note two things. First, in both schems we have the same probability for hitting the jack pot and hence winning. Second, the probability that we DON'T hit the jack pot is higher in the first scheme. This means that it is more likely for us to lose our money under the first scheme (and hence casino winning) than the second scheme. Therefore, the casino would prefer the first scheme better since although their risk for losing is the same in either scheme, their chance for winning is higher under the first scheme. ______________________________________________________________________________ (p.71, #5) Note that in this problem we are concerned about "your birthday," and not "a birthday." a. Define, E = the event that no one else in the class has MY birthday. Then, P(at least one other student has my birthday) = 1 - P(E). But, P(E) = (364/365)^(n-1) since any one of the other n-1 students in the class can have any birthday except for mine. Note that in this case there could be other birthday matches amongst the rest of the class. The only event that we care about is no match with my birthday. So, each of the other n-1 students in the class can pick any day of the year for their birthday, except for mine. Namely, there are 364 choices for each of the remaining n-1 students. b. Set 1 - P(E) \geq 1/2, where \geq means "greater than or equal to." Thus, P(E) \leq 1/2 => (364/365)^(n-1) \leq 1/2 Now, solve this inequality using the logarithmic function. Note that log(364/365) is a negative number, so when you divide the inequality by this neg number the direction of the inequality has to be reveresed. c. The difference is clear. In this problem we looked for no match with my birthday, but in the birthday problem we looked for no match in the class. ______________________________________________________________________________