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For one roll of a certain die, the probability of rolling a

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Senior Manager
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For one roll of a certain die, the probability of rolling a [#permalink] New post 11 Sep 2011, 00:38
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

91% (01:44) correct 9% (01:47) wrong based on 11 sessions
For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled
4 times, which of the following is the probability that the outcome will be a two at
least 3 times?
[strike](A) (1/6)4
(8) 2(1/6i + (1/6)4
(C) 3{1/6)3(5/6) + (1/6)4
(O) 4{1/6)3{5/6) + (1/6)4
(E) 6{1/6)3{5/6) + (1/6)4[/strike]

Corrected options:
(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3(5/6) + (1/6)^4
(D) 4(1/6)^3(5/6) + (1/6)^4
(E) 6(1/6)^3(5/6) + (1/6)^4
[Reveal] Spoiler: OA

Last edited by bb on 10 Oct 2011, 11:47, edited 3 times in total.
Options Corrected
Senior Manager
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Re: For one roll of a certain die [#permalink] New post 11 Sep 2011, 00:41
should there be 5 cases for this one ??


Outcomes:
2 - 2 - 2 - not a 2

2 - 2 - 2 - 2

2 -not a 2 - 2- 2

not a 2 - 2 - 2 - 2

2 - 2 - not a 2 - 2


???
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Re: For one roll of a certain die [#permalink] New post 11 Sep 2011, 06:03
siddhans wrote:
Tried searching for this one but couldnt find it...

For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled
4 times, which of the following is the probability that the outcome will be a two at
least 3 times?
(A) (1/6)4
(8) 2(1/6i + (1/6)4
(C) 3{1/6)3(5/6) + (1/6)4
(O) 4{1/6)3{5/6) + (1/6)4
(E) 6{1/6)3{5/6) + (1/6)4


The following is the question with correct format. Please format your questions properly. The math symbol to denote power is "^".

For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3(5/6) + (1/6)^4
(D) 4(1/6)^3(5/6) + (1/6)^4
(E) 6(1/6)^3(5/6) + (1/6)^4

*********************************************

Sol:
This is the question in which you make use of binomial distribution.

What is the probability that the outcome will be a two AT LEAST 3 times:
Outcome 2 EXACTLY 3 times
+
Outcome 2 EXACTLY 4 times

Outcome 2 EXACTLY 3 times
=(\frac{1}{6})^3*(1-\frac{1}{6})^1*\frac{4!}{3!}=4*(\frac{1}{6})^3(\frac{5}{6})^1

Outcome 2 EXACTLY 4 times
=(\frac{1}{6})^4*(1-\frac{1}{6})^0=(\frac{1}{6})^4

Outcome 2 EXACTLY 3 times = Outcome 2 EXACTLY 3 times + Outcome 2 EXACTLY 4 times

=4(\frac{1}{6})^3(\frac{5}{6})+(\frac{1}{6})^4

Ans: "D"

******************************************************
In its simplest form:

Answer=\frac{7}{2*6^3}=\frac{7}{432}

Read this:
permutation-combination-and-probabilities-14706.html#p89947
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Re: For one roll of a certain die [#permalink] New post 11 Sep 2011, 06:31
this is based on binomial distribution .


= ncr (p^r)(1-p)^(n-r)

Probability of getting two atleast 3 times out of 4 attempts
= p(two appearing exactly 3 times) + p( two appearing exactly 4 times)
= 4c3 * (1/6)^3 * (5/6) + 4c4 (1/6)4


Answer is D.
Re: For one roll of a certain die   [#permalink] 11 Sep 2011, 06:31
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