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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]  26 Sep 2010, 08:38
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65% (02:34) correct 35% (00:15) wrong based on 20 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.

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

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

Hope it's clear.
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Re: Coin Toss [#permalink]  26 Sep 2010, 08:56
Expert's post
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]  26 Sep 2010, 19:31
1
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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
Manager
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Kudos [?]: 234 [1] , given: 104

Re: Coin Toss [#permalink]  27 Sep 2010, 00:28
1
KUDOS
Thank you guys! I felt this question was interesting.
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Re: Coin Toss [#permalink]  27 Sep 2010, 12:00
Awesome question, it tests important fundamentals
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Re: Coin Toss [#permalink]  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 ($$\frac{5!}{2!3!}$$).

Hope it's clear.

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