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4 dices are thrown at the same time. What is the probability

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Joined: 06 Sep 2013
Posts: 1661
Location: United States
Concentration: Finance
GMAT 1: 710 Q48 V39
WE: Corporate Finance (Investment Banking)
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Kudos [?]: 164 [0], given: 268

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Re: Really tough [#permalink] New post 20 Feb 2014, 11:21
Bunuel wrote:
roshanaslam wrote:
4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9
B. 5/9
C. 11/18
D. 7/9
E. none of the above


I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}.

Answer: B.


If we have 4C2 aren't we assuming that the other two also have to be the same number? Or does it work for 2 the same number, the other two different?

I thought that we had to use 4!/2!1!1! because the last two terms were different.

Could anybody please clarify
Thanks
Cheers
J
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Re: Really tough [#permalink] New post 20 Feb 2014, 23:39
Expert's post
jlgdr wrote:
Bunuel wrote:
roshanaslam wrote:
4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9
B. 5/9
C. 11/18
D. 7/9
E. none of the above


I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}.

Answer: B.


If we have 4C2 aren't we assuming that the other two also have to be the same number? Or does it work for 2 the same number, the other two different?

I thought that we had to use 4!/2!1!1! because the last two terms were different.

Could anybody please clarify
Thanks
Cheers
J


We have 4 dice (A, B, C, D).

C^2_4 is the ways to select which two dice out of 4, will provide the same face: (A, B), (A, C), (A, D), (B, C), (B, D), or (C, D) . Next, these two dice can take 6 values, the remaining two 5 and 4, respectively. So, the # of favorable outcomes is C^2_4*6*5*4.

Hope it's clear.
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Re: Really tough   [#permalink] 20 Feb 2014, 23:39
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