Math 340 Homework Set 9 ____________________________________________________________________________ 1. (p.128, #3) In this problem you need to recall the basic def for the prob of an event A: P(A) = (# of ways that A can occur) / (total # of possible outcomes). a. A = the event that the first player holds all aces. i. Total # of ways that we can give 13 cards to a player (without replacement and without order) = 52 choose 13. ii. Now for A to occur, we first have to choose all 4 aces in "4 choose 4" ways and then choose 9 more cards, in "48 choose 9" ways, to be handed to the first player. Thus, P(A) = (ii) / (i). b. B = the event that she holds the ace of hearts. i. We are looking for P(A|B). ii. From the def of cond prob we know that P(A|B) = P(AB) / P(B). Hence we need to calculate P(AB) and P(B). iii. For B to occur first we need to hand her the ace of hearts in exactly one possible way. Then, we hand her another 12 cards in any of the "51 choose 12" ways. Hence, P(B) = (1 times "51 choose 12") / ("52 choose 13") iv. Now for AB to occur we first hand her the ace of hearts in one possible way, then hand her three more aces in "3 choose 3" ways, and then we hand her 9 more cards in "48 choose 9" ways. Thus, P(AB) = (1 times "3 choose 3" times "48 choose 9") / ("52 choose 13"). v. Hence, (ii)-(iv) now provide the soln. Other parts of this problem are obtained using the same approach. c. To get the answer in the form that appears in the back of the book, we can think about the problem in the following way: She can choose the 4 aces in (4 choose 4) ways, and she can choose the remaining 9 cards for her hand in (48 choose 9) ways. Thus, the total number of ways that she can have a hand with all four aces is (4 choose 4) (48 choose 9) Now, the total number hands is (the number of ways that a 13 card hand can be chosen from the full deck) - (the number of ways that a 13 card hand can be chosen from the remaining 48 cards, since none of the aces are available to us). This is (52 choose 13) - (48 choose 13). The ratio gives the answer in the back of the book. Note that there is another way to think about this problem. That is to consider P(all aces | at least one)=P(all aces AND at least one)/P(at least onn) Now the two probabilities in the top and the bottom are not too had to calculate. The answer will be equivalent to the answer in the back of the book, but it will be set up differently. _____________________________________________________________________________ 2. (p.128, #5) Let, X = the no. of successes in n trials, and Y = the proportion of successes in n trials, namely, Y=X/n. Then, the distribution of X is B(n,p), mu=np, and sigma=\sqrt(npq). Hence, the distribution of Y has a mean mu=p, and stand dev. sigma=\sqrt(pq/n). In this problem p=55%, and we are solving for n so that with probability 99% more than p_1 = 50% of the voters favor candidate A. In other words, find n such that 0.99 = P(Y \geq 0.5). Now, we can use an appropriate normal distribution to find n: 0.99 = P(Y \geq 0.5) = P(Z > z) = 1 - P(Z < z). Hence, P(Z < z) = 1 - 0.99 = 0.01 => z = -2.33 (from table) But, z = (p_1 - p)/sigma => -2.33 = (0.5-0.55)/sqrt[(0.55)(0.45)/n] If we solve this equation for n we get n=537.46 or about 538. ADDITIONAL COMMENTS: Let X = B(n, p). Then X represents the total number of successes in n trials. The expected value of X is mu = n p and the S.D. of X is sigma = sqrt(n p q). Define a new r.v. P-hat = X / n. The P-hat represents the proportion of successes in a sequence of n Berboulli trials. Since P-hat is obtained from X by dividing by n, we can get the mean and the S.D. of P-hat by dividing the mean and the SD of X by n. Hence the expected value and the SD of P-hat are: P-hat = X / n mu = np / n = p sigma = sqrt(n p q) / n = sqrt(pq/n) Furthermore, since for large n X may be approximated by a normal dist N(np, sqrt(npq)), we may also conclude that for large n P-hat may be approximated by a normal dist N(p, sqrt9pq/n)). Now, in this problem we are goven that "the chance for the majority of voters to be in favor of candidate A is 99%." Namely, 0.99 = P(P-hat > 0.50). (We use 0.50 since they said majority.) Since P-hat may be approximated by a corresponding normal dist, the above probability is approximately equal to: 0.99 = P(Z > z), where z is the z-score for 0.50. Since p is known to be 0.55, we get the following equation for z: z = (0.50 - 0.55) / sqrt[(0.55)(1-0.55)/n] From the table in Appendix 5, using the right tail probability 99%, we get z = -2.33. Substitute this value of z in the above equation and solve for n. REMARK: THE RANDOM VARIABLE P-HAT IN THIS PROBLEM IS CALLED THE SAMPLE PROPORTION R.V. ______________________________________________________________________________ 3. (p.128, #7) The first thing that you need to note is that there are 80 balls and we are drawing 4 in various schemes. Thus, regardless of the scheme, total number of ways that we can draw 4 balls from a box of 80 balls is: total number of possible outcomes = "80 choose 4" This is the term that has to go in the denominator of all prob calculations. a. Let A = the event that all four balls are black. Then, P(A) = ("50 choose 4") / ("80 choose 4"), since we are drawing balls without replacement or order. b. B = the event that exactly three balls are black. Then, the red ball can be drawn on the 1st, or ..., or 4th draw. Hence P(B) = ( 4 (50)_3 30 / 4!) / ("80 choose 4") = ("50 choose 3") ("30 choose 1") / ("80 choose 4") Note that we have to divide by 4! since we are drawing balls without order. c. C = the event that the first red ball appears on the last draw. Then, this is like above except that we know when the red ball is drawn. Thus, P(C) = ( (50)_3 30 / 4! ) / ("80 choose 4").