Problem Solving

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. 16/27

OA is d

Source- Kaplan Practice Test

pareekbharat86 wrote: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. 16/27

P(exactly n times) = P(one way) * total possible ways.

Let A = a result of A and N = a result of not A.
Since P(A) = 1/3, P(N) = 2/3.

P(one way):
Over the course of 5 experiments, ONE WAY to get exactly 2 A's is AANNN.
P(A is yielded by the 1st experiment) = 1/3.
P(A is yielded by the 2nd experiment) = 1/3.
P(N is yielded by the 3rd experiment) = 2/3.
P(N is yielded by the 4th experiment) = 2/3.
P(N is yielded by the 5th experiment) = 2/3.
Since we want all of these events to happen, we MULTIPLY:
1/3 * 1/3 * 2/3 * 2/3 * 2/3 = 8/243.

Total possible ways:
AANNN is only ONE WAY to get exactly 2 A's.
Now we must account for ALL OF THE WAYS to get exactly 2 A's.
Any arrangement of the letters AANNN represents one way to get exactly 2 A's and 3 N's.
Thus, to account for ALL OF THE WAYS to get exactly 2 A's and 3 N's, the result above must be multiplied by the number of ways to arrange the letters AANNN.
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical A's and by 3! to account for the two identical N's:
5!/(2!3!) = 10.

Multiplying the results above, we get:
P(exactly 2 A's) = 10 * 8/243 = 80/243.

The correct answer is D.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html

A's probability = 1/3
Not Happening probability = 2/3

Exactly twice = [(2/3)^3 * (1/3)^2]* 5!/2!3! = [8/243] * 10 = 80/243

Answer [spoiler]{D}[/spoiler]

GMATGuruNY wrote:

pareekbharat86 wrote: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. 16/27

P(exactly n times) = P(one way) * total possible ways.

Let A = a result of A and N = a result of not A.
Since P(A) = 1/3, P(N) = 2/3.

P(one way):
Over the course of 5 experiments, ONE WAY to get exactly 2 A's is AANNN.
P(A is yielded by the 1st experiment) = 1/3.
P(A is yielded by the 2nd experiment) = 1/3.
P(N is yielded by the 3rd experiment) = 2/3.
P(N is yielded by the 4th experiment) = 2/3.
P(N is yielded by the 5th experiment) = 2/3.
Since we want all of these events to happen, we MULTIPLY:
1/3 * 1/3 * 2/3 * 2/3 * 2/3 = 8/243.

Total possible ways:
AANNN is only ONE WAY to get exactly 2 A's.
Now we must account for ALL OF THE WAYS to get exactly 2 A's.
Any arrangement of the letters AANNN represents one way to get exactly 2 A's and 3 N's.
Thus, to account for ALL OF THE WAYS to get exactly 2 A's and 3 N's, the result above must be multiplied by the number of ways to arrange the letters AANNN.
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical A's and by 3! to account for the two identical N's:
5!/(2!3!) = 10.

Multiplying the results above, we get:
P(exactly 2 A's) = 10 * 8/243 = 80/243.

The correct answer is D.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html

Hi Mitch,

Thanks for your response.

The calculation of 8/243 seemed simple. I could also get upto there. However, I am not too comfortable with why we went on to calculate 5C2 here.

Why is there a need to consider combinations here. The question simply asks if a random experiment is conducted 5 times, what would be the prob. to have A occurring exactly twice. Why do we need to repeat it for various combinations?