For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) \((\frac{1}{6})^4\)

(B) \(2(\frac{1}{6})^3 + (\frac{1}{6})^4\)

(C) \(3(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4\)

(D) \(4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4\)

(E) \(6(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4\)


Kudos for a correct solution.
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Solution -
Below are the chances of getting a two at least 3 times after die rolled 4 times:

One chance is not favorable, three chances are favorable -
NSSS - (5/6)*(1/6)^3
SNSS - (1/6)*(5/6)*(1/6)^2
SSNS - (1/6)^2*(5/6)*(1/6)
SSSN - (1/6)^3*(5/6)

All the chances are favorable -

YYYY - (1/6)^4

Sum up all the above possibilities = \(4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4\)

ANS D

Ah, dang. I see my mistake. The answer is indeed D.

Thanks,

Option D

3 two's
Probability=4!/3!) *(1/6)*(1/6)*(1/6)*(5/6)
4 two'2
Probability=4!/4! * (1/6)*(1/6)*(1/6)*(1/6)

Sum of these two probabilities gives answer D

MANHATTAN GMAT OFFICIAL SOLUTION:

Unfortunately, you cannot easily use the 1 - x trick here, so you must express the probability directly. You must regard the desired outcome in two separate parts: first, rolling a two exactly 4 times, and second, rolling a two exactly 3 times out of four attempts. First, the probability of rolling a two exactly 4 times is (1/6)(1/6)(1/6)(1/6) = (1/6)^4.

Next, if you roll a two exactly 3 times out of 4 attempts, then on exactly one of those attempts, you do not roll a two. Hence, the probability of rolling a two exactly 3 times out of 4 attempts is the sum of the following four probabilities:



Notice that there are 4 rearrangements of 3 “Twos” and 1 “Not a two”. In other words, you have to count as separate outcomes the 4 different positions in which the “Not a two” roll occurs: first, second, third, or fourth.

There is no way to roll a two exactly 4 times AND exactly 3 times, so you can now just add up these probabilities. Thus, the desired probability is 4(1/6)^3(5/6) + (1/6)^4.

Answer: D.

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