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Gerry has a hand of 5 playing cards, drawn from the same standard 52

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Bunuel
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Gerry has a hand of 5 playing cards, drawn from the same standard 52 [#permalink] New post   26 Oct 2021, 02:45
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49% (02:48) correct 51% (02:56) wrong based on 61 sessions
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Gerry has a hand of 5 playing cards, drawn from the same standard 52-card deck. Exactly four of the cards are clubs, and exactly two of the cards are sevens. Two cards are selected at random from Gerry’s hand. What is the probability that both of the cards are clubs, and at least one of the cards is a seven?

A. 1/5
B. 3/10
C. 2/5
D. 3/5
E. 7/10
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chetan2u
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Re: Gerry has a hand of 5 playing cards, drawn from the same standard 52 [#permalink] New post   29 Oct 2021, 07:50
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Bunuel wrote:
Gerry has a hand of 5 playing cards, drawn from the same standard 52-card deck. Exactly four of the cards are clubs, and exactly two of the cards are sevens. Two cards are selected at random from Gerry’s hand. What is the probability that both of the cards are clubs, and at least one of the cards is a seven?

A. 1/5
B. 3/10
C. 2/5
D. 3/5
E. 7/10



So, what is the information we have.
5 cards that have 4 clubs and 2 sevens .
Thus surely one of the sevens is of club and what we have is a 7 of other than club, a 7 of club and three other cards of club.

Selecting two cards out of 5: 5C2 = 10 ways.

Selection with restrictions: both club and one 7.
Thus, one of the two is 7 of club and other can be chosen out of remaining three clubs in 3C1 or 3 ways.

P= \(\frac{3}{10}\)


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Re: Gerry has a hand of 5 playing cards, drawn from the same standard 52 [#permalink] New post   28 Apr 2023, 10:35
chetan2u wrote:
Bunuel wrote:
Gerry has a hand of 5 playing cards, drawn from the same standard 52-card deck. Exactly four of the cards are clubs, and exactly two of the cards are sevens. Two cards are selected at random from Gerry’s hand. What is the probability that both of the cards are clubs, and at least one of the cards is a seven?

A. 1/5
B. 3/10
C. 2/5
D. 3/5
E. 7/10



So, what is the information we have.
5 cards that have 4 clubs and 2 sevens .
Thus surely one of the sevens is of club and what we have is a 7 of other than club, a 7 of club and three other cards of club.

Selecting two cards out of 5: 5C2 = 10 ways.

Selection with restrictions: both club and one 7.
Thus, one of the two is 7 of club and other can be chosen out of remaining three clubs in 3C1 or 3 ways.

P= \(\frac{3}{10}\)


B


I didn't understand about how we are getting 3C1.

I understand there is 3/5 possibility of getting both clubs (i.e. 4/5 * 3/4). But I don't understand how to proceed after that. Please help. Thanks.
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Re: Gerry has a hand of 5 playing cards, drawn from the same standard 52 [#permalink] New post   28 Apr 2023, 10:52
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fdfd97 wrote:
chetan2u wrote:
Bunuel wrote:
Gerry has a hand of 5 playing cards, drawn from the same standard 52-card deck. Exactly four of the cards are clubs, and exactly two of the cards are sevens. Two cards are selected at random from Gerry’s hand. What is the probability that both of the cards are clubs, and at least one of the cards is a seven?

A. 1/5
B. 3/10
C. 2/5
D. 3/5
E. 7/10



So, what is the information we have.
5 cards that have 4 clubs and 2 sevens .
Thus surely one of the sevens is of club and what we have is a 7 of other than club, a 7 of club and three other cards of club.

Selecting two cards out of 5: 5C2 = 10 ways.

Selection with restrictions: both club and one 7.
Thus, one of the two is 7 of club and other can be chosen out of remaining three clubs in 3C1 or 3 ways.

P= \(\frac{3}{10}\)


B


I didn't understand about how we are getting 3C1.

I understand there is 3/5 possibility of getting both clubs (i.e. 4/5 * 3/4). But I don't understand how to proceed after that. Please help. Thanks.


fdfd97

There are three clubs other than the seven-of-clubs...you want one of the three...3C1.

For a question like this with very manageable numbers, I'm a fan of writing out the elements to help with visualization and choosing the correct numbers.

There are five cards:
seven-of-clubs
blank1-of-clubs
blank2-of-clubs
blank3-of-clubs
blank-of-nonclubs

In order to win, we need the seven-of-clubs plus one of the other clubs...there are three ways to do that...out of a total of ten possible pairs of cards. 3/10

Answer choice B.
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Re: Gerry has a hand of 5 playing cards, drawn from the same standard 52 [#permalink]
28 Apr 2023, 10:52
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