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Manager  Status: Keep fighting!
Joined: 31 Jul 2010
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When a random experiment is conducted, the probability that  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 69% (01:50) correct 31% (02:20) wrong based on 208 sessions

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 53796
Re: Coin Toss  [#permalink]

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5
1
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]

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5
1
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.
##### General Discussion
Manager  Joined: 22 Aug 2008
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Re: Coin Toss  [#permalink]

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1
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  Status: Keep fighting!
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Re: Coin Toss  [#permalink]

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1
Thank you guys! I felt this question was interesting.
Joined: 31 Dec 1969
Location: Russian Federation
Concentration: Entrepreneurship, International Business
WE: Supply Chain Management (Energy and Utilities)
Re: Coin Toss  [#permalink]

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Awesome question, it tests important fundamentals
Intern  Joined: 19 Aug 2010
Posts: 8
Re: Coin Toss  [#permalink]

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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!
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Re: When a random experiment is conducted, the probability that  [#permalink]

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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?

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]

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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?

A. 5/243
B. 25/243
C. 64/243
D. 80/243
E. 16/17

We can let y = event A occurs and n = event A does not occur.

Thus, the probability that event A occurs two times followed by three non-occurrences of event A is:

P(y-y-n-n-n) = 1/3 x 1/3 x 2/3 x 2/3 x 2/3 = 8/243

However, we also have to consider that the outcome of 2 Ys and 3 Ns can occur in other orderings. For example, we can have (n-y-n-n-y-n) or (n-n-n-y-y), and so forth. The number of ways for these rearrangements to occur can be calculated by using the formula for permutations of indistinguishable objects: 5!/(3! x 2!) = 10.

Thus, the probability is 8/243 x 10 = 80/243.

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Intern  G
Joined: 01 Jan 2016
Posts: 29
Re: When a random experiment is conducted, the probability that  [#permalink]

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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.

Is applying the binomial formula correct in this problem? i.e, (5c2)(1/3)^2(2/3)^3.. the results are same, but is it the right approach?
Manager  B
Joined: 11 Jun 2017
Posts: 69
Re: When a random experiment is conducted, the probability that  [#permalink]

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P(event A occurs) = 1/3
P (event A does not occur) = 2/3
P(event A will occur exactly 2 times)= 1/3*1/3*(2/3)^2 = 8/243
Now two experiments can occur in 5!/2!*3! ways as 2 events are similar and the rest are similar too.
= 8/243*10 = 80/243 (D)
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Re: When a random experiment is conducted, the probability that  [#permalink]

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# When a random experiment is conducted, the probability that

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