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When a random experiment is conducted, the probability that [#permalink]
26 Sep 2010, 08:38

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Question Stats:

65% (02:24) correct
35% (00:47) wrong based on 40 sessions

When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). 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

took more time than I hoped for in the tests (Kaplan). Don't forget to KUDOS if you like the question.

When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?

5/243 25/243 64/243 80/243 16/17

took more time than I hoped for in the tests (Kaplan). Don't forget to KUDOS if you like the question.

Tx.

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

1. number of ways to choose 2 occasion in which A would occur is 5c2=10 2. out of 5 times, the probability of occurring A is (1/3)(1/3)(1-2/3)(1-2/3)(1-2/3)=8/243

When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?

5/243 25/243 64/243 80/243 16/17

took more time than I hoped for in the tests (Kaplan). Don't forget to KUDOS if you like the question.

Tx.

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

Re: When a random experiment is conducted, the probability that [#permalink]
18 Jul 2015, 05:22

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Re: When a random experiment is conducted, the probability that [#permalink]
18 Jul 2015, 05:41

Expert's post

hemanthp wrote:

When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). 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

took more time than I hoped for in the tests (Kaplan). Don't forget to KUDOS if you like the question.

Tx.

Probability of The event to occur = 1/3 i.e. Probability of The event to NOT occur = 1-(1/3) = (2/3)

Let Occurrence is represented by 'O' and Non-Occurrence is represented by 'N'

Then the total ways of two occurrences out of 5 trials is same as arrangement of 5 letters 'OONNN'

Arrangement of 'OONNN' = 5C2 or 5!/(2!)(3!) = 10

i.e. The probability of two occurrences and 3 non-occurrences for one arrangement = (1/3)(1/3)(2/3)(2/3)(2/3) = (1/3)^2*(2/3)^3

i.e. The probability of two occurrences and 3 non-occurrences for all the arrangement = 5C2*(1/3)^2*(2/3)^3 = 10*(1/3)^2*(2/3)^3 = 80/243

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