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Unformatted text preview: Solutions to Homework 6, Math 55 Section 5.1 6. There are 4 aces, 13 hearts, and one ace of hearts; thus, by inclusionexclusion, 4 + 13 1 = 16 cards are aces or hearts. Therefore, the probability a random card is an ace or a heart is 16 52 = 4 13 . 10. There are ( 52 5 ) total combinations of cards, out of which ( 50 3 ) contain the two of diamonds and the three of spades. Therefore, the probability a hand has these two cards is ( 50 3 ) / ( 52 5 ) = 19600 2598960 = 5 663 . 16. For each suit, there are ( 13 5 ) flushes in that suit, so there are 4 ( 13 5 ) hands which are flushes. Thus, the probability of a flush is 4 ( 13 5 ) / ( 52 5 ) = 5148 2598960 = 13 16660 . 22. Out of the 100 positive integers not exceeding 100, b 100 3 c = 33 are divisible by 3; thus, the probability such an integer is divisible by 3 is 33 100 . 28. There are ( 80 7 ) possible tickets; out of these, ( 11 7 ) are winners. Therefore, the probability of winning is ( 11 7 ) / ( 80 7 ) = 3 / 28879240. Alternately, if we treat the ticket as being fixed, then there are ( 80 11 ) ways for the lottery commissioners to choose the 11 winning numbers. Out of these, there are ( 73 4 ) ways for them to choose your 7 numbers and 4 others. Therefore, the probability of winning is ( 73 4 ) / ( 80 11 ) = 3 / 28879240. 36. When two dice are rolled, there are 5 ways to get a total of 8; thus, the probability of getting 8 is 5 36 . When three dice are rolled, the number of ways to get a total of 8 is the same as the number of solutions to x 1 + x 2 + x 3 = 5, where x i is one less than the number on the i th die. This is ( 5+3 1 5 ) = ( 7 2 ) = 21, so the probability of getting a total of 8 is 21 6 3 = 7 72 . (Note that since the sum was 5, we didnt have to worry about the fact that each x i 5.) Since 5 36 > 7 72 , getting a total of 8 is more likely with two dice than with three....
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 Spring '08
 STRAIN
 Math

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