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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:
70% (02:15) correct
30% (01:17) wrong based on 54 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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Re: When a random experiment is conducted, the probability that
[#permalink]
18 Jul 2015, 05:41
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