Quiz9Soln

# Quiz9Soln - C(26 4 Â C(26 5 = 983 411 000 ways 2 Four cards...

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Math 17C, Spring 2009 Instructor: Timothy Lewis TA: Tamara Schlichter (Section A01) QUIZ 9 - SOLUTIONS Name: Student ID: Show all work for credit. 20 pts possible. 1. (10 pts) A standard deck of cards consists of 52 cards: 26 black and 26 red. In how many ways can four red and ﬁve black cards be selected from a standard deck of cards if cards are drawn without replacement? The four red cards can be drawn in C (26 , 4) = 14 , 950 ways. The ﬁve black cards can be drawn in C (26 , 5) = 65 , 780 ways. Therefore the total number of ways =
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Unformatted text preview: C (26 , 4) Â· C (26 , 5) = 983 , 411 , 000 ways. 2. Four cards are drawn at random without replacement from a standard deck of 52 cards. A) (5 pts) What is the probability of no ace? The probability of obtaining no aces at all = ( 48 4 ) ( 52 4 ) = 0 . 7187. B) (5 pts) What is the probability of at least one ace? The probability of obtaining at least one ace = 1-Probability of obtaining no aces at all. So the probability of obtaining at least one ace = 1-( 48 4 ) ( 52 4 ) = 0 . 2813....
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## This note was uploaded on 10/27/2009 for the course MATH 17C taught by Professor Lewis during the Spring '09 term at UC Davis.

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