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

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20% (00:51) wrong based on 45 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?

--------------------------------------------------------------------------------------- If you think you can,you can If you think you can't,you are right.

Last edited by Bunuel on 08 Jun 2012, 05:49, edited 2 times in total.

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 _________________

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

.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 (\(\frac{5!}{2!3!}\)).

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