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For each player's turn in a certain board game, a card is [#permalink]
06 May 2012, 09:29

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This post was BOOKMARKED

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E

Difficulty:

65% (hard)

Question Stats:

61% (03:26) correct
39% (02:51) wrong based on 112 sessions

For each player's turn in a certain board game, a card is drawn. 3/4 of the cards in the deck are marked with a circle, and the 1/4 remaining cards are marked with a square. If five players draw a card and then return it to the deck, what is the probability that at least four of the cards drawn are marked with a square?

A. (1/4)^3 B. 5(1/4)^3 C. 3/4(1/4)^4 D. 2/3(1/4)^4 E. (1/4)^5

Re: For each player's turn in a certain board game, a card is [#permalink]
06 May 2012, 09:41

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Galiya wrote:

For each player's turn in a certain board game, a card is drawn. 3/4 of the cards in the deck are marked with a circle, and the 1/4 remaining cards are marked with a square. If five players draw a card and then return it to the deck, what is the probability that at least four of the cards drawn are marked with a square?

A. (1/4)^3 B. 5(1/4)^3 C. 3/4(1/4)^4 D. 2/3(1/4)^4 E. (1/4)^5

At least four of the cards drawn are marked with a square means 4 or all 5 cards are marked with a square.

\(P=P(SSSSC)+P(SSSSS)=\frac{5!}{4!}*(\frac{1}{4})^4*\frac{3}{4}+(\frac{1}{4})^5=\frac{1}{4^3}\), we are multiplying first case by \(\frac{5!}{4!}\), since SSSSC can occur in several ways: SSSSC, SSSCS, SSCSS, ... Notice that the number of occurrences of SSSSC basically is the number of arrangements of 5 letters SSSSC out of which 4 S's are identical, so it's \(\frac{5!}{4!}\).

Re: For each player's turn in a certain board game, a card is [#permalink]
02 Dec 2013, 04:43

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Re: For each player's turn in a certain board game, a card is [#permalink]
03 May 2015, 02:12

Hello from the GMAT Club BumpBot!

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For each player's turn in a certain board game, a card is [#permalink]
02 Sep 2015, 14:01

Bunuel wrote:

Galiya wrote:

For each player's turn in a certain board game, a card is drawn. 3/4 of the cards in the deck are marked with a circle, and the 1/4 remaining cards are marked with a square. If five players draw a card and then return it to the deck, what is the probability that at least four of the cards drawn are marked with a square?

A. (1/4)^3 B. 5(1/4)^3 C. 3/4(1/4)^4 D. 2/3(1/4)^4 E. (1/4)^5

At least four of the cards drawn are marked with a square means 4 or all 5 cards are marked with a square.

\(P=P(SSSSC)+P(SSSSS)=\frac{5!}{4!}*(\frac{1}{4})^4*\frac{3}{4}+(\frac{1}{4})^5=\frac{1}{4^3}\), we are multiplying first case by \(\frac{5!}{4!}\), since SSSSC can occur in several ways: SSSSC, SSSCS, SSCSS, ... Notice that the number of occurrences of SSSSC basically is the number of arrangements of 5 letters SSSSC out of which 4 S's are identical, so it's \(\frac{5!}{4!}\).

Answer: A.

Hope it's clear.

Hi Bunuel,

Can we do this by Binomial Distribution as well ?

Thanks, Gaurav

gmatclubot

For each player's turn in a certain board game, a card is
[#permalink]
02 Sep 2015, 14:01

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