Explanation 1) Formula: Probability of the event P ( A ) = n ( A ) n ( S ) . Where, n ( A ) is the number of elements in the set of outcomes in event A . n ( S ) is the number of elements in sample space . P ( A ′ ) is the complement of the event A . (i.e.) P ( A ′ ) = 1 − P ( A ) 2) Calculation: Given, Number of cast in the deck n ( S ) = 52 Number of face cars in the deck = 12 Number of non-face cards in the deck = 40 Let N be the event of draw a non-face card without replacement from the deck Set of outcomes in the event N = { number of non-face card } n ( N ) = 40 Then, P ( N ) be the probability that we draw a non-face card without replacement from the deck. P ( N ) = n ( N ) n ( S ) P ( N ) = 40 52 P ( N ) = 10 13 Again draw a non-face card without replacement from the deck after a non-face card. Now, number of cards in the deck after draws a card = n ( S ) − 1 n ( S ) − 1 = 52 − 1 n ( S ) − 1 = 51 Number of non-face cards in the deck after draw a non-face card = n ( N ) − 1 n ( N ) − 1 = 40 − 1 = 39 Let P ( N / a non-face card was drawn ) be the probability that we draw a non-face card from the deck after a card drawn without replacement

6th Edition

ISBN: 9780134506470

Author: Pirnot

Publisher: PEARSON

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Chapter 13.3, Problem 21E

To determine

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The probability that we draw at least one face card of drawing 2 cards without replacement from the deck.

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Chapter 13.3, Problem 20E

Chapter 13.3, Problem 22E

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The probability that we draw at least one face card of drawing 2 cards without replacement from the deck.

Solution Summary: The author explains that the probability of drawing a non-face card without replacement from the deck is 0.412.

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Chapter 13 Solutions

To find:

The probability that we draw at least one face card of drawing 2 cards without replacement from the deck.