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__

$$\Pr of\ rolling\ a\ two\ \left[P\left(T\right)\right]=\frac{1}{6}$$
$$\Pr of\ not\ rolling\ a\ two\ \left[P\left(N\right)\right]=1-\frac{1}{6}=\frac{5}{6}$$
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.