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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]

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New post 26 Sep 2010, 09:38
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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.
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Re: Coin Toss  [#permalink]

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New post 26 Sep 2010, 09:48
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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 (\(\frac{5!}{2!3!}\)).

Answer: D.

Hope it's clear.
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Re: Coin Toss  [#permalink]

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New post 26 Sep 2010, 09:56
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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]

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New post 26 Sep 2010, 20:31
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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]

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New post 27 Sep 2010, 01:28
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Thank you guys! I felt this question was interesting.
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Re: Coin Toss  [#permalink]

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New post 27 Sep 2010, 13:00
Awesome question, it tests important fundamentals
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Re: Coin Toss  [#permalink]

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New post 28 Sep 2010, 03:18
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!
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Re: When a random experiment is conducted, the probability that  [#permalink]

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New post 18 Jul 2015, 06:41
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

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

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New post 20 Jul 2017, 16:48
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.

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

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New post 16 Sep 2017, 02:38
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.



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

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New post 16 Sep 2017, 06:54
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 &nbs [#permalink] 16 Sep 2017, 06:54
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