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A standard deck of 52 cards is shuffled, and 2 cards...
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Updated on: 24 Jan 2019, 01:03
Updated on: 24 Jan 2019, 01:03
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A standard deck of 52 cards is shuffled, and 2 cards are drawn, one after another without replacement. A standard deck has 4 aces. It also has 12 face cards: 4 kings, 4 queens, and 4 jacks. For the first draw all of the aces are removed. For the second draw the aces are returned to the deck and all of the face cards are removed. What is the probability that the first draw is a face card and the second draw is an ace?
A) \(\frac{2}{221}\)
B) \(\frac{3}{169}\)
C) \(\frac{1}{40}\)
D) \(\frac{12}{221}\)
E) \(\frac{4}{13}\)
A) \(\frac{2}{221}\)
B) \(\frac{3}{169}\)
C) \(\frac{1}{40}\)
D) \(\frac{12}{221}\)
E) \(\frac{4}{13}\)
Originally posted by CAMANISHPARMAR on 23 Jan 2019, 23:48.
Last edited by Carcass on 24 Jan 2019, 01:03, edited 1 time in total.
Last edited by Carcass on 24 Jan 2019, 01:03, edited 1 time in total.
Edited by Carcass
Re: A standard deck of 52 cards is shuffled, and 2 cards...
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24 Jan 2019, 01:08
24 Jan 2019, 01:08
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Expert Reply
The first step is the aces out so we do have 48 cards and 12 faced cards.
\(\frac{12}{48} = \frac{1}{4}\)
For the second step we do have 40 cards and 4 aces so we do have \(\frac{4}{40}\) \(= \frac{1}{10}\)
\(\frac{1}{4} * \frac{1}{10} = \frac{1}{40}\)
C is the answer
\(\frac{12}{48} = \frac{1}{4}\)
For the second step we do have 40 cards and 4 aces so we do have \(\frac{4}{40}\) \(= \frac{1}{10}\)
\(\frac{1}{4} * \frac{1}{10} = \frac{1}{40}\)
C is the answer
Re: A standard deck of 52 cards is shuffled, and 2 cards...
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05 Sep 2021, 08:48
05 Sep 2021, 08:48
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