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When a random experiment is conducted, the probability that

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Manager
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When a random experiment is conducted, the probability that [#permalink]

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29 Jan 2010, 11:26
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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, 06:49, edited 2 times in total.
Edited the question and added the OA
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29 Jan 2010, 11:34
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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29 Jan 2010, 20:42
What am I missing here?

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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08 Jun 2012, 06:41
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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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08 Jun 2012, 06:50
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 ($$\frac{5!}{2!3!}$$).

Hope it's clear.
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When a random experiment is conducted, the probability that [#permalink]

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05 Oct 2014, 04:09
1
KUDOS
one case is:
1/3*1/3*2/3*2/3*2/3=2^3/3^5

we have 5!/2!*3!=10 such cases

so, 2^3*10/3^5=80/243

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Re: When a random experiment is conducted, the probability that [#permalink]

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18 Feb 2016, 23:47
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Re: When a random experiment is conducted, the probability that   [#permalink] 18 Feb 2016, 23:47
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