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

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02 Nov 2012, 13:43
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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 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. 1/4^4
C. 5*(1/4^3)
D. 1/4^5
E. (3/2)*(1/4^4)
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Nov 2012, 13:56, edited 1 time in total.
Renamed the topic and added OA.
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02 Nov 2012, 14:03
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clarkkent0610 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 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. 1/4^4
C. 5*(1/4^3)
D. 1/4^5
E. (3/2)*(1/4^4)

The probability of circle is 3/4 and the probability of square is 1-3/4=1/4.

We want the probability that out of 5 cards drawn 4 OR 5 are squares.

The probability of 4 squares or the probability of SSSSC, is $$P(SSSSC)=\frac{5!}{4!}*(\frac{1}{4})^4*(\frac{3}{4})=\frac{15}{4^5}$$. We are multiplying by $$\frac{5!}{4!}=5$$ since SSSSC scenario can occur in 5 ways: SSSSC, SSSCS, SSCSS, SCSSS, and CSSSS (number of premutations of 5 letters SSSSC out of which 4 S's are identcal);

The probability of 5 squares or the probability of SSSSS, is simply $$P(SSSSS)=(\frac{1}{4})^5$$.

Therefore the overall probability is $$\frac{15}{4^5}+\frac{1}{4^5}=\frac{1}{4^3}$$.

P.S. Please indicate OA for PS problems.
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28 Mar 2016, 04:49
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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 remaining cards are marked with a square. If 5 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) $$3/2( 1/4 )^{4}$$
(E) $$( 1/4)^{4}$$

[Reveal] Spoiler:
A
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28 Jul 2014, 08:24
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16 Sep 2015, 02:38
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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28 Mar 2016, 04:54
AnnaKathpal 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 remaining cards are marked with a square. If 5 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) $$3/2( 1/4 )^{4}$$
(E) $$( 1/4)^{4}$$

[Reveal] Spoiler:
A

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

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30 Oct 2016, 11:25
Hi Bunuel,

I have got a doubt with the usage of Permutation or Combination in such types of questions. For the following problem:http://gmatclub.com/forum/if-jesse-flips-a-coin-seven-times-in-a-row-what-is-the-probability-th-211560.html

In the above problem bernoulli's theoreom is used: 7C5(There are 7C5 ways that the heads can be placed in 7 times) but in this problem permutation is used: (number of premutations of 5 letters SSSSC out of which 4 S's are identical).

Are the usage of these both interchangeable? Do they mean the same? If yes, can you please explain me how?
Re: For each players turn in a certain board game, a card is   [#permalink] 30 Oct 2016, 11:25
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