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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 26 Sep 2010, 08:38
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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.
[Reveal] Spoiler: OA
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Re: Coin Toss [#permalink] New post 26 Sep 2010, 08:48
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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?

    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!}).

Answer: D.

Hope it's clear.
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Re: Coin Toss [#permalink] New post 26 Sep 2010, 08:56
This can be broken down into two parts:

1. The ways in which two of these events can occur - Combination
2. The probability of the event occurring.

For part 1: We have 2 events out of 5 events and hence the ways you can choose two out of five events is: \frac{5!}{2! * 3!} = 10

For part 2: Probability of correct coin toss = \frac{1}{3} and probability of not happening = 1 - \frac{1}{3} = \frac{2}{3}

Hence required probability = \frac{1}{3} * \frac{1}{3} * \frac{2}{3}* \frac{2}{3}* \frac{2}{3} * 10 = \frac{80}{243}

Hope this helps.
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Re: Coin Toss [#permalink] New post 26 Sep 2010, 19:31
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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

So the final probability is 10*(8/243)=80/243
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Re: Coin Toss [#permalink] New post 27 Sep 2010, 00:28
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Thank you guys! I felt this question was interesting.
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Re: Coin Toss [#permalink] New post 27 Sep 2010, 12:00
Awesome question, it tests important fundamentals
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Re: Coin Toss [#permalink] New post 28 Sep 2010, 02:18
Bunuel wrote:
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?

    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!}).

Answer: D.

Hope it's clear.


Great explanation.Thanks!
Re: Coin Toss   [#permalink] 28 Sep 2010, 02:18
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