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# For each player's turn in a certain board game, a card is

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For each player's turn in a certain board game, a card is [#permalink]

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06 May 2012, 10:29
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Question Stats:

61% (03:24) correct 39% (02:51) wrong based on 122 sessions

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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
[Reveal] Spoiler: OA
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Re: For each player's turn in a certain board game, a card is [#permalink]

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06 May 2012, 10:41
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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 $$\frac{5!}{4!}$$.

Hope it's clear.
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Re: For each player's turn in a certain board game, a card is [#permalink]

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06 May 2012, 10:56
great!Thank you so much,Bunuel!
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Re: For each player's turn in a certain board game, a card is [#permalink]

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02 Dec 2013, 05:43
Hello from the GMAT Club BumpBot!

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

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02 Dec 2013, 06:26
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Bunuel,
You have mentioned P(SSSC)+P(SSSSS)
it should be P(SSSSC)+ P(SSSSS),right?
Regards.
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Re: For each player's turn in a certain board game, a card is [#permalink]

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02 Dec 2013, 08:15
Expert's post
Bunuel,
You have mentioned P(SSSC)+P(SSSSS)
it should be P(SSSSC)+ P(SSSSS),right?
Regards.

Yes, one S was missing. Edited. Thank you.
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Re: For each player's turn in a certain board game, a card is [#permalink]

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03 May 2015, 03:12
Hello from the GMAT Club BumpBot!

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For each player's turn in a certain board game, a card is [#permalink]

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02 Sep 2015, 15:01
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 $$\frac{5!}{4!}$$.

Hope it's clear.

Hi Bunuel,

Can we do this by Binomial Distribution as well ?

Thanks,
Gaurav
For each player's turn in a certain board game, a card is   [#permalink] 02 Sep 2015, 15:01
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