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When a random experiment is conducted, the probability

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When a random experiment is conducted, the probability

by M7MBA » Fri Mar 16, 2018 4:12 am
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

The OA is the option D.

What is the best way to solve this PS question? I couldn't find a short way to solve it. <i class="em em-sob"></i>
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by GMATGuruNY » Fri Mar 16, 2018 4:50 am
M7MBA 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. 16/17
P(exactly n times) = P(one way) * (total number of ways).

Let Y = the probability that the event occurs and N = the probability that the event does NOT occur.

P(one way):
One way to get exactly two Y's over 5 experiments:
YYNNN.
P(YYNNN) = 1/3 * 1/3 * 2/3 * 2/3 * 2/3 = 8/243.

Total possible ways:
YYNNN is only ONE WAY to get exactly two Y's.
Now we must account for ALL OF THE WAYS to get exactly two Y's.
Any arrangement of the letters YYNNN represents one way to get exactly two Y's.
Thus, to account for ALL OF THE WAYS to get exactly two Y's, the result above must be multiplied by the number of ways to arrange the letters YYNNN.
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 2 identical Y's and by 3! to account for the 3 identical N's:
5!/(2!3!) = 10.

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

The correct answer is D.

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by swerve » Fri Mar 16, 2018 12:44 pm
Hi M7MBA,

I solved this PS question as follow,

1. The number of ways to chose 2 occasions 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. Option D.

Regards!
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by Scott@TargetTestPrep » Mon Mar 19, 2018 3:34 pm
M7MBA 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. 16/17
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) 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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by deloitte247 » Thu Mar 29, 2018 1:01 pm
The probability that even A occur = 1/3
The probability that even A will not occur= $$1-\frac{1}{3}=\frac{\left(3-1\right)}{3}=\frac{2}{3}$$
$$Let\ the\ probability\ that\ event\ A\ occur=E=\frac{1}{3}$$
$$Let\ the\ probability\ that\ event\ A\ will\ not\ occur=F=\frac{2}{3}$$ $$The\ total\ ways\ of\ 2\ occurrences\ out\ of\ 5\ trials=E,\ E,\ F,\ F,\ F$$
$$The\ arrangement\ goes\ as\ follow$$
$$The\ permutation\ of\ 5\ letters\ \left(E,E,F,F,F\right)\ out\ of\ which\ 2E's\ and\ 3F's\ are\ identical\ or\ alike=\frac{5!}{2!3!}$$
$$The\ probability\ of\ 2\ occurrences\ \left[E^2=\left(\frac{1}{3}\right)^{^{^2}}\right]\ and\ 3\ non-occurrences\ \left[p^3=\left(\frac{2}{3}\right)^3\right]\ for\ all\ the\ arrangement=$$ $$\frac{5!}{2!3!}\cdot\left(\frac{1}{3}\right)^2\cdot\left(\frac{2}{3}\right)^3$$
$$=\left[\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot3\cdot2\cdot1}\right]\cdot\left[\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{2}{3}\right]$$
$$=\left[\frac{120}{12}\right]\cdot\left[\frac{8}{243}\right]=\frac{80}{243}$$
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