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math_403-01_spring_2009_final_exam_solutions

math_403-01_spring_2009_final_exam_solutions - St Louis...

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St. Louis University Spring 2009 Solutions: Final Exam Math 403–01 Prof. G. Marks 1. (15 pts.) A box contains 100 dice. Of these, 99 dice are cubes, and the remaining die is a tetrahedron (a pyramid with a triangular base, having four faces). Each cubical die is labeled with the integers 1 through 6 , and these numbers occur with equal probability when the die is rolled. The tetrahedral die is labeled with the integers 1 through 4 , and these numbers occur with equal probability when the die is rolled. Someone randomly selects two dice from the box and rolls them. The sum of the two numbers appearing on the selected dice is 9 . What is the probability that one of the two dice selected was tetrahedral? Solution. Let A be the event that two cubes are selected from the box; let B be the event that one cube and one tetrahedron are selected. Let N be the event that 9 is the total on the two selected dice. Then Prob( A ) = parenleftbigg 99 2 parenrightbigg parenleftbigg 100 2 parenrightbigg = 49 50 , Prob( B ) = 1 - Prob( A ) = 1 50 , Prob( N | A ) = 1 9 , Prob( N | B ) = 1 12 (since there are four ways to roll 9 on two cubical dice out of 36 ways to roll anything, and there are two ways to roll 9 on a cube and tetrahedron out of 24 ways to roll anything). Given that 9 was rolled, the conditional probability that one of the selected dice was tetrahedral is Prob( B | N ) = Prob( B N ) Prob( N ) = Prob( N | B ) · Prob( B ) Prob( N ) = Prob( N | B ) · Prob( B ) Prob( N | A ) · Prob( A ) + Prob( N | B ) · Prob( B ) (by Bayes’s Theorem) = 1 12 · 1 50 1 9 · 49 50 + 1 12 · 1 50 = 3 199 . 2. (16 pts.) The card game bridge is played by dealing a standard deck of 52 cards * to four players, known as North, South, East, and West, so that each player receives 13 cards. If South has been dealt precisely five spades, what is the probability that North has been dealt three or more spades? (Assume the deck was shuffled before the deal so that every arrangement of the 52 cards is equally likely.) Solution. There are parenleftbigg 39 13 parenrightbigg different hands that North can be dealt, since this is the number of ways of selecting 13 cards from among the 39 cards not in South’s hand. The number of North hands that contain precisely k spades (for k ∈ { 0 , 1 , 2 , . . ., 8 } ) is parenleftbigg 8 k parenrightbigg · parenleftbigg 31 13 - k parenrightbigg , since there are ( 8 k ) ways to select k spades for North’s hand from among the eight spades not in South’s hand, and ( 31 13 - k ) ways to select 13 - k non-spades for North’s hand from among the 31 non-spades not in South’s hand. * A standard deck consists of four suits, clubs, diamonds, hearts, and spades; each suit contains cards of 13 denominations. 1

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Thus, given that South has five spades, the conditional probability that North has k spades is parenleftbigg 8 k parenrightbigg · parenleftbigg 31 13 - k parenrightbigg parenleftbigg 39 13 parenrightbigg for k ∈ { 0 , 1 , 2 , . . ., 8 } .
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math_403-01_spring_2009_final_exam_solutions - St Louis...

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