Explanation Given: You are to draw one card from a deck. If it is an ace, you win $ 10 ; if it is a face card, you win $ 1 ; otherwise, you lose $0.50. Approach: The probability of an event E is, P ( E ) = n ( E ) n ( S ) … ( 1 ) Here, n ( E ) denotes the favorable outcome, n ( S ) denotes the total outcome. The formula for the expected value of an event is, E = a 1 p 1 + a 2 p 2 + ⋅ ⋅ ⋅ + a n p n … ( 2 ) Here, E is the expected value. Calculation: Let X , Y , and Z be the event that the card is an ace, a face card, and a card other than an ace and a face card respectively. From the given information, n ( X ) = 4 , n ( S ) = 52 . Substitute 4 for n ( X ) , and 52 for n ( S ) in equation ( 1 ) P ( X ) = 4 52 = 1 13 Now, n ( Y ) = 12 , n ( S ) = 52 . Substitute 4 for n ( Y ) , 52 for n ( S ) in equation ( 1 ) P ( Y ) = 12 52 = 3 13 Now, n ( Z ) = 36 , n ( S ) = 52

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ISBN: 8220100548512

Author: Watson

Publisher: CENGAGE L

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Chapter 14.CT, Problem 15CT

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The expected value of the game.

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EBK ALGEBRA AND TRIGONOMETRY

Ch. 14.1 - CONCEPTS The Fundamental Counting Principle says...Ch. 14.1 - CONCEPTS The number of ways of arranging r objects...Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E Ch. 14.1 - Prob. 10E

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The expected value of the game.

Solution Summary: The author explains that the expected value of the game is 0.65.

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To find:

The expected value of the game.