practice2-solutions

# practice2-solutions - UNIVERSITY OF VICTORIA MATHEMATICS...

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Unformatted text preview: UNIVERSITY OF VICTORIA MATHEMATICS 151, SECTION A02, FALL 2008 PRACTICE TEST 2 SOLUTIONS Question 1. A casino operates a game where the player flips a coin and rolls a 6 sided die. If they get a heads and roll a 4 , 5 or 6, they win \$20. If they get a tails and roll a 1 or a 3, they win \$36. Otherwise, they win nothing. If the casino wishes to make an average a profit of \$0 . 96 per player, how much should the casino charge to play the game? (A) \$0 (B) \$3 (C) \$6 (D) \$9 (E) \$12 (F) \$15 (G) \$18 (H) \$21 (I) \$24 (J) \$27 Solution: The player wins \$20 with a probaility of 1 4 , and \$36 with a probability of 1 6 . Therefore the expected value is (\$20) 1 4 + (\$36) 1 6 = \$5 + \$6 = \$11 . If they wish to make a profit of \$0 . 96, they need to charge \$11 . 96. So the answer is (E). Question 2. You have a drawer with 10 blue socks, 8 red socks, and 6 grey socks. If you draw 3 socks, what is the probability that at least 2 of them are blue? (A) . 05 (B) . 15 (C) . 25 (D) . 35 (E) . 45 (F) . 55 (G) . 65 (H) . 75 (I) . 85 (J) . 95 Solution: If X counts the number of blue socks drawn without replacement, we have N = 24 socks, K = 10 red socks, and n = 3 socks drawn. We wish to consider the chance of getting 2 blue, plus the chance of getting 3 blue socks. Using the formula for the hypergeometric distribution we have P ( X ≥ 2) = ( 10 2 )( 14 1 ) ( 24 3 ) + ( 10 3 )( 14 ) ( 24 3 ) = 630 2 , 024 120 2 , 024 = 750 2 , 024 ∼ . 37055336 ....
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practice2-solutions - UNIVERSITY OF VICTORIA MATHEMATICS...

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