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805+ (Hard)|   Probability|      
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Bunuel
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Gerry has a hand of 5 playing cards, drawn from the same standard 52 26 Oct 2021, 02:45
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95% (hard)

Question Stats:

51% (02:43) correct 49% (02:43) wrong based on 166 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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B
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Gerry has a hand of 5 playing cards, drawn from the same standard 52 29 Oct 2021, 07:50
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Bunuel
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
Show more


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
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Gerry has a hand of 5 playing cards, drawn from the same standard 52 28 Apr 2023, 10:35
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chetan2u
Bunuel
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
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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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Gerry has a hand of 5 playing cards, drawn from the same standard 52 28 Apr 2023, 10:52
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fdfd97
chetan2u
Bunuel
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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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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Gerry has a hand of 5 playing cards, drawn from the same standard 52 25 Jun 2024, 07:44
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Bunuel
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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Gerry has a hand of 5 playing cards, drawn from the same standard 52 13 May 2025, 02:58
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We can arrange Club and Club Seven in 3 * 1 = 3 ways
Total ways = 5C2 = 10 ways

Answer = 3/10 = 0.3
Bunuel
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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