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When a random experiment is conducted, the probability that

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When a random experiment is conducted, the probability that [#permalink] New post 29 Jan 2010, 10:26
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B
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E

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82% (01:51) correct 18% (01:15) wrong based on 17 sessions
When a random experiment is conducted, the probability that event A occurs is 1/3. If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?

A. 5/243
B. 25/243
C. 64/243
D. 80/243
E. 16/17
[Reveal] Spoiler: OA

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Last edited by Bunuel on 08 Jun 2012, 05:49, edited 2 times in total.
Edited the question and added the OA
Expert Post
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Re: Probability [#permalink] New post 29 Jan 2010, 10:34
Expert's post
mojorising800 wrote:
When a random experiment is conducted, the probability that event A occurs is ????. If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?

You need to add information about the probability that even A occurs
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Re: Probability [#permalink] New post 29 Jan 2010, 19:42
What am I missing here? :roll:

Lets say getting A is represented by W and not getting A by L.
prob of A = 3/10
prob of not getting A= 7/10
Out of 5 times, getting 2 As can be done in 5C2 = 10 ways.

So prob of getting 2 As = 10*3/10 * 3/10 * 7/10 * 7/10 * 7/10 = 3087 / 10000
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Re: Probability [#permalink] New post 08 Jun 2012, 05:41
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I think this should be

[.3 * .3 * .7 * .7 * .7 ] * 5!/ 3! 2!=

.03087 * 10 = .3087 how ever this does not match any given options
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Re: Probability [#permalink] New post 08 Jun 2012, 05:50
Expert's post
stne wrote:
I think this should be

[.3 * .3 * .7 * .7 * .7 ] * 5!/ 3! 2!=

.03087 * 10 = .3087 how ever this does not match any given options


You are right. There was a typo in the stem, 0.3 instead of 1/3. Typo edited. Thank you.

Complete solution:
When a random experiment is conducted, the probability that event A occurs is 1/3. If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?
A. 5/243
B. 25/243
C. 64/243
D. 80/243
E. 16/17

The probability that event A occurs is \frac{1}{3};
The probability that event A will not occur is 1-\frac{1}{3}=\frac{2}{3}.

We want to calculate the probability of event YYNNN (Y-occurs, N-does not occur).

P(YYNNN)=\frac{5!}{2!3!}*(\frac{1}{3})^2*(\frac{2}{3})^3=\frac{80}{243}, we are multiplying by \frac{5!}{2!3!} as event YYNNN can happen in # of times: YYNNN, YNYNN, NNNYY, ... basically the # of permutations of 5 letters YYNNN out of which 2 Y's and 3 N's are identical (\frac{5!}{2!3!}).

Answer: D.

Hope it's clear.
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Re: Probability   [#permalink] 08 Jun 2012, 05:50
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