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# For each player's turn in a certain board game, a card is drawn. 3/4

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Joined: 28 Oct 2012
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For each player's turn in a certain board game, a card is drawn. 3/4  [#permalink]

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Updated on: 29 Jan 2018, 19:45
5
20
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Difficulty:

95% (hard)

Question Stats:

49% (02:25) correct 51% (02:28) wrong based on 219 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 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)

Originally posted by clarkkent0610 on 02 Nov 2012, 13:43.
Last edited by Bunuel on 29 Jan 2018, 19:45, edited 2 times in total.
Renamed the topic and added OA.
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Joined: 02 Sep 2009
Posts: 62501
Re: For each player's turn in a certain board game, a card is drawn. 3/4  [#permalink]

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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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Re: For each player's turn in a certain board game, a card is drawn. 3/4  [#permalink]

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29 Jan 2018, 19:25
Hi Bunuel,
I did it this way, please correct me. idk what did I miss out
Quote:
what is the probability that at least four of the cards drawn are marked with a square?

- 4 squares = 5C4 * (1/4)^4 * (3/4)^1
- 5 squares = 5C5 * (1/4)^5 * (3/4)^0
Total prob = (21/4) * (1/4)^4
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Posts: 62501
Re: For each player's turn in a certain board game, a card is drawn. 3/4  [#permalink]

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29 Jan 2018, 19:51
lichting wrote:
Hi Bunuel,
I did it this way, please correct me. idk what did I miss out
Quote:
what is the probability that at least four of the cards drawn are marked with a square?

- 4 squares = 5C4 * (1/4)^4 * (3/4)^1
- 5 squares = 5C5 * (1/4)^5 * (3/4)^0
Total prob = (21/4) * (1/4)^4

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

$$P(SSSSC)=\frac{5!}{4!}*(\frac{1}{4})^4*(\frac{3}{4})=\frac{15}{4^5}$$.

$$P(SSSSS)=(\frac{1}{4})^5$$.

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

This is explained here: https://gmatclub.com/forum/for-each-pla ... l#p1139227

In the highlighted part in your solution it's not clear how you got the numbers there and why are multiplying instead of adding.
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Re: For each player's turn in a certain board game, a card is drawn. 3/4  [#permalink]

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30 Mar 2019, 10:32
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Re: For each player's turn in a certain board game, a card is drawn. 3/4   [#permalink] 30 Mar 2019, 10:32
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