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PS - Good probablity question

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xALIx
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PS - Good probablity question [#permalink] New post   Updated on: 17 Jun 2008, 16:02
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Tricky question, thought I'd share with the group. Took me 3 mins to solve :-(

A fair, six-sided die, with sides numbered one through six, is rolled 5 times. What is the probability that on exactly 3 rolls the number of dots showing is greater than 4?
A) 1/243
B) 10/243
C) 1/27
D) 40/243
E) 80/243

edit:
I mistyped, it should say "What is the probability that on exactly..."



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Originally posted by xALIx on 17 Jun 2008, 10:32.
Last edited by xALIx on 17 Jun 2008, 16:02, edited 2 times in total.
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Re: PS - Good probablity question [#permalink] New post   17 Jun 2008, 10:53
Please rephrase the question. It's too confusing to be able to answer properly.
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Re: PS - Good probablity question [#permalink] New post   17 Jun 2008, 11:09
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I think answer is (D)

In order to have more than 4 on exactly 3 rolls, you have to get 5 or 6 on 3 rolls (chance this happens for 1 roll: 2/6) and 1,2,3 or 4 on the 2 others (chance this happens for 1 roll: 4/6).

Once you chose which of the 3 rolls will bear the 5 or 6, you have (2/6)^3 * (4/6)^2 chances it happens

This is 4/243

And there are 10 choices of the 3 rolls on 5 (unordered choices).

Therefore answer is 40/243.
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Re: PS - Good probablity question [#permalink] New post   17 Jun 2008, 11:24
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Agree with Oski

\(p=C^5_3*(\frac{2}{6})^3*(\frac{4}{6})^2=\frac{5!*2^2}{3!*2!*3^5}=\frac{40}{243}\)
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Re: PS - Good probablity question [#permalink] New post   17 Jun 2008, 16:04
x-ALI-x wrote:
Tricky question, thought I'd share with the group. Took me 3 mins to solve :-(

A fair, six-sided die, with sides numbered one through six, is rolled 5 times. What is the probability that on exactly 3 rolls the number of dots showing is greater than 4?
A) 1/243
B) 10/243
C) 1/27
D) 40/243
E) 80/243

edit:
I mistyped, it should say "What is the probability that on exactly..."



For this question, the answer is 40/243, the key is including the combinations so 4/243*10
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Re: PS - Good probablity question [#permalink] New post   17 Jun 2008, 16:05
I'll state the question differently, for similar practice:

A fair, six-sided die, with sides numbered one through six, is rolled 5 times. What is the probability that on exactly 3 rolls the number of dots showing is no greater than 4?
A) 1/243
B) 10/243
C) 1/27
D) 40/243
E) 80/243

btw, the answer is different from the one above
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Re: PS - Good probablity question [#permalink] New post   18 Jun 2008, 00:46
Same reasoning gives :
\(p=C^5_3*(\frac{4}{6})^3*(\frac{2}{6})^2=\frac{5!}{3!*2!}*\frac{2^3}{3^5}=\frac{80}{243}\)



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: PS - Good probablity question [#permalink]
18 Jun 2008, 00:46
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