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There are 26 black cards and we want 4 of them ne c 26

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Unformatted text preview: Solution: There are 26 red cards and we want 0 of them. There are 26 black cards and we want 4 of them. n(E ) = C (26, 0)C (26, 4) n(S ) = C (54, 4) P (E ) = C (26,0)C (26,4) C (52,4) ≈ .0552 (E) one of the cards is a club, two of the cards are spades, and the other card is a diamond? Solution: There are 13 clubs and we want 1 of them. There are 13 spades and we want 2 of them. There are 13 diamonds and we want 1 of them. n(E ) = C (13, 1)C (13, 2)C (13, 1) n(S ) = C (52, 4) P (E ) = C (13,1)C (13,2)C (13,1) C (52,4) ≈ .0487 2 Problem 3 A quiz consists of 5 true-or-false questions. If a student guesses at every answer, what is the probability that he or she will answer (A) exactly 3 of the questions correctly? Solution: n(E ) = C (5, 3) n(S ) = 2 × 2 × 2 × 2 × 2 = 25 P (E ) = C (5,3) 25 = .3125 (B) at least 1 of the questions correctly? Solution: Notice the “at least 1.” That means that you will use the complement rule. If E is the event that at least 1 is correct, then E c is the event that none is correct. n(E c ) = C (5, 0) n(S ) = 25 P (E c ) = C (5,0) 25 P (E ) = 1 − P (E c ) = 1 − C (5,0) 25 ≈ .9688 (C) the first and last question correctly? Solution: n( E ) = 1 × 2 × 2 × 2 × 1 = 2 3 n(S ) = 25 P (E ) = 23 25 = .25 (D) all of the questions correctly? Solution: n(E ) = C (5, 5) n(S ) = 25 P (E ) = C (5,5) 25 = .03125 3 Problem 4 A customer from Cavallaro’s Fruit Stand picks a sample of 5 oranges at random from a crate containing 75 oranges, of which 6 are rotten. What is the probability that the sample contains (A) exactly 2 rotten oranges? Solution: There are 6 rotten oranges and we want 2 of them. There are 69 good oranges and we want 3 of them (since we need a total of 5 oranges.) n(E ) = C (6, 2)C (69, 3) n(S ) = C (75, 5) P (E ) = C (6,2)C (69,3) C (75,5) ≈ .0455 (B) 3 or 4 rotten oranges? Solution: We want 3 rotten, 2 good OR 4 rotten, 1 good. We will add the number of ways we can have 3 rotten, 2 good to the number of ways we can have 4 rotten, 1 good. n(E ) = C (6, 3)C (69, 2) + C (6, 4)C (69,...
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