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# PS - Good probablity question

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PS - Good probablity question [#permalink]

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17 Jun 2008, 09:32
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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..."

Last edited by xALIx on 17 Jun 2008, 15:02, edited 2 times in total.
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Re: PS - Good probablity question [#permalink]

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17 Jun 2008, 09:53
Please rephrase the question. It's too confusing to be able to answer properly.
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Re: PS - Good probablity question [#permalink]

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17 Jun 2008, 10:09
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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]

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17 Jun 2008, 10:24
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]

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17 Jun 2008, 15:04
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]

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17 Jun 2008, 15:05
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]

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17 Jun 2008, 23:46
Same reasoning gives :
$$p=C^5_3*(\frac{4}{6})^3*(\frac{2}{6})^2=\frac{5!}{3!*2!}*\frac{2^3}{3^5}=\frac{80}{243}$$
Re: PS - Good probablity question   [#permalink] 17 Jun 2008, 23:46
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