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For one roll of a certain die, the probability of rolling a

by AAPL » Tue Aug 27, 2019 6:06 am
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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. \((1/6)^4\)
B. \(2(1/6)^3+(1/6)^4\)
C. \(3(1/6)^3(5/6)+(1/6)^4\)
D. \(4(1/6)^3(5/6)+(1/6)^4\)
E. \(6(1/6)^3(5/6)+(1/6)^4\)

OA D
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by Jay@ManhattanReview » Tue Aug 27, 2019 10:09 pm
AAPL wrote:Manhattan Prep

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. \((1/6)^4\)
B. \(2(1/6)^3+(1/6)^4\)
C. \(3(1/6)^3(5/6)+(1/6)^4\)
D. \(4(1/6)^3(5/6)+(1/6)^4\)
E. \(6(1/6)^3(5/6)+(1/6)^4\)

OA D
Probability of getting a 'Two' = 1/6.

Probability that the outcome will be a 'Two' at least 3 times

= Probability that the outcome will be a 'Two' 3 times + Probability that the outcome will be a 'Two' 4 times
= 4C3*(1/6)^3*(5/6) + 4C4*(1/6)^4
= 4*(1/6)^3*(5/6) + (1/6)^4

The correct answer: D

Hope this helps!

-Jay
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by swerve » Wed Aug 28, 2019 12:45 pm
AAPL wrote:Manhattan Prep

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. \((1/6)^4\)
B. \(2(1/6)^3+(1/6)^4\)
C. \(3(1/6)^3(5/6)+(1/6)^4\)
D. \(4(1/6)^3(5/6)+(1/6)^4\)
E. \(6(1/6)^3(5/6)+(1/6)^4\)

OA 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__
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by deloitte247 » Thu Aug 29, 2019 8:09 pm
$$\Pr of\ rolling\ a\ two\ \left[P\left(T\right)\right]=\frac{1}{6}$$
$$\Pr of\ not\ rolling\ a\ two\ \left[P\left(N\right)\right]=1-\frac{1}{6}=\frac{5}{6}$$
Pr of rolling a two at least three times = Pr of rolling two 3 times + Pr of rolling 2 four times since the max rolling is 4 times.
$$\Pr of\ TTTN=\frac{4!}{3!}\cdot\left(\frac{1}{6}\right)^{^3}\cdot\frac{5}{6}$$
$$\Pr of\ TTTN=\frac{4\cdot3!}{3!}\cdot\left(\frac{1}{6}\right)^{^3}\cdot\frac{5}{6}$$
$$=4\cdot\left(\frac{1}{6}\right)^{^3}\cdot\frac{5}{6}$$
$$\Pr\ of\ TTTT=\frac{4!}{4!}\cdot\left(\frac{1}{6}\right)^{^4}=\left(\frac{1}{6}\right)^{^4}$$
$$\Pr\ of\ rolling\ a\ two\ and\ at\ least\ 3\ timesTTTT=\left[4\cdot\left(\frac{1}{6}\right)^{^3}\cdot\frac{5}{6}\right]+\left(\frac{1}{6}\right)^{^4}$$
$$=4\cdot\left(\frac{1}{6}\right)^{^3}\cdot\frac{5}{6}+\left(\frac{1}{6}\right)^{^4}$$

In conclusion, option D is our correct choice of pick.
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by Scott@TargetTestPrep » Mon Sep 02, 2019 5:54 pm
AAPL wrote:Manhattan Prep

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. \((1/6)^4\)
B. \(2(1/6)^3+(1/6)^4\)
C. \(3(1/6)^3(5/6)+(1/6)^4\)
D. \(4(1/6)^3(5/6)+(1/6)^4\)
E. \(6(1/6)^3(5/6)+(1/6)^4\)

OA D
Let Y = rolling a 2 and N = not rolling a 2. The probability of Y is â…™, and the probability of N is â…š.

We need to determine the probability of Y-Y-Y-Y or Y-Y-Y-N

Since the probability of Y on each roll is 1/6, the probability of Y-Y-Y-Y is (1/6)^4.
For Y-Y-Y-N, since the probability of each roll is 1/6, the probability of Y-Y-Y-N is (1/6)^3(5/6), and since Y-Y-Y-N can be arranged in 4!/3! = 4 ways, the probability is 4(1/6)^3(5/6).

Thus, the overall probability is (1/6)^4 + 4(1/6)^3(5/6).

Answer: D

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