When a random experiment is conducted, the probability that

Question

When a random experiment is conducted, the probability that event A occurs is 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

whiplash2411 · Accepted Answer

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.

Bunuel · Answer

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.

geturdream · Answer

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

aarthis2 · Answer

Great explanation.Thanks!

Temurkhon · Answer

one case is:
1/3*1/3*2/3*2/3*2/3=2^3/3^5

we have 5!/2!*3!=10 such cases

so, 2^3*10/3^5=80/243

D

GMATinsight · Answer

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

JeffTargetTestPrep · Answer

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

Poorvasha · Answer

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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