### Section 4.4

• 2. The probability is 1/6, approx. = 0.17, since there are 6 equally likely outcomes.
• 4. April has 30 days. The answer is therefore 30/366 , approx. = 0.082.
• 6. There are 16 cards that qualify as being an ace or a heart, so the answer is 16/52, approx. = 0.31. This can also be computed from Theorem 2 as 4/52+13/52- 1/52.
• 8. From Example 5, there are C(52,5) possible poker hands, and we assume by symmetry that they are all equally likely. In order to solve this problem, we need to compute the no. of poker hands that contain the ace of hearts. There is no choice about choosing the ace of hearts. To form the rest of the hand we need to choose 4 cards from the 51 remaining cards, so there are C(51,4) hands containing the ace of hearts. Therefore, the answer is the ratio: C(51,4)/C(52,5)=5/52, approx. = 9.6 %. The problem can also be done by subtracting from 1 the answer to Exercise 9, since a hand contains the ace of hearts if and only if it is not the case that it does not contain the ace of hearts.
• 10. This is similar to Exercise 8. We need to compute the number of poker hands that contain the 2 of diamonds and the 3 of spades. There is no choice about choosing these 2 cards. To form the rest of the hand, we need to choose 3 cards from the 50 remaining cards, so there are C(50,3) hands containing these 2 specific cards. Therefore the answer to the question is the ratio C(50,3)?C(52,5)=5/663, approx. = 0.0075.
• 12. There are 4 ways to specify the ace. Once the ace is chosen for the hand, there are C(48,4) ways to choose the nonaces for the reamaining 4 cards. Therefore there are 4C(48,4) hands with exactly 1 ace. Since there are C(52,5) equally likely hands, the answer is the ratio 4C(48,4)/C(52,5), approx. = 0.30.
• 14. We saw in example 5 that there are C(52,5) = 2,598,960 different hands, and we assume by symmetry that they are equally likely. We need to count the number of hands that have 5 different kinds (ranks). There are C(13,5) ways to choose the kinds. For each card, there are then 4 ways to choose the suit. Therefore, there are C(13,5).4^5=1,317,888 ways to choose the hand. Thus the probability is 1317888/2598960, approx. = 0.51.
• 16. Of the C(52,5) = 2,598,960 hands, 4.C(13,5)=5148 are flushes, since we can specify a flush by choosing a suit and then choosing 5 cards from that suit. Therefore the answer is 5148/2598960=33/16660, approx. = 0.0020.
• 18. There are clearly only 10.4 = 40 straight flushes, since all we get to specify for a straight flush is the starting (lowest) kind in the straight (anything from ace up to ten) and the suit. Therefore, the answer is 40/C(52,5)= 40/2598960=1/64974.
• 20. There are 4 royal flushes, 1 in each suit. Therefore the answer is 4/C(52,5)=4/2598960=1/649740.
• 22. There are (upper bound)100/3 = 33 multiples of 3 among the integers from 1 to 100 (inclusive), so the answer is 33/100=0.33.
• 24. In each case, the numbers are chosen from the integers from 1 to n, then there are C(n,6) possible entries, only one of which is the winning one, so the answer is 1/C(n,6). a) 1/C(30,6), approx. = 1.7 x 10^-6. b) 1/C(36,6), approx. = 5.1 x 10^-7. c) 1/C(42,6), approx. = 1.9 x 10^-7. d) 1/C(48,6), approx. = 8.1 x 10^-8.
• 26. In each case, if the numbers are chosem from the integers from 1 to n, then there are C(n,6) possible entries. If we wish to avoid all the winning numbers, then we must make out choice from the n-6 nonwinning numbers, and this can be done in C(n-6,6) ways. Therefore, since the winning numbers are picked at random, the probability is C(n-6,6)/C(n,6). a) C(34,6)/C(40,6), approx. = 0.35. b) C(42,6)/C(48,6), approx. = 0.43. c) C(50,6)/C(56,6), approx. = 0.49. d) C(58,6)/C(64,6), approx. = 0.54.
• 28. We need to compute the number of ways for the Pennsylvania lottery commission to select its 11 numbers, and the number of ways for it to select its 11 numbers so as to contain the 7 numbers that we chose. For the former, the number is clearly C(80,11) . For the latter, the commission must select 4 more numbers besides the ones we chose, from the 80-7 = 73 other numbers, so there are C(73,4) ways to do this. Therefore, the probability that we win is the ratio C(73,4)/C(80,11), which works out to 3/28879240, or about 1 chance in 10 million (1.04 x 10^-7). The same answer can be obtained by counting things in the other direction: the number of ways for us to choose 7 of the commission's predestined 11 numbers divided by the number of ways for us to pick 7 numbers. This gives C(11,7)/C(80,7), which has the same value as before.
• 30. In order to specifya winning ticket, we must choose 5 of the 6 numbers to match (C(6,5) = 6 ways to do so) and one number from the among the remaining 34 numbers not to match (C(34,1)=34 ways to do so). Therefore there are 6.34 = 204 winning tickets. Since there are C(40,6)=3,838,380 tickets in all, the answer is 204/3838380, approx. = 5.3 x 10^-5, or about 1 chance in 19,000.