For one roll of a certain number cube with six faces, numbered 1 throu

Total rolls = 4 times

Dice = 6 sided

Probability of a "2" on every roll = 1/6

Probability of not a "2" but any of the other 5 numbers = 1 - 1/6 = 5/6

Probability of "2" at least 3 times means :

a. 2 three times

b. 2 four times

Case 1 : T T T N (T = two and N = Not two)

\(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{5}{6} * \frac{4!}{3!}\)

We multiply by 4! because permutations/total arrangements of 4 different items can be done in 4! ways

We divide by 3! because we have 3 identical items.

Therefore, 4*(1/6)^4 * 5/6

Case 2 : T T T T (All twos)

\((1/6)^4\)


Add Case I and II

\(4*(1/6)^4 * 5/6\) + \((1/6)^4\)

Given that For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6 and the cube is rolled 4 times and We need to find which of the following is the probability that the outcome will be a two at least 3 times?

As we are rolling a cube 4 times => Number of cases = \(6^4\) = 1296

Now, P(getting a two at least 3 times) = P(Getting 2 exactly 3 times) + P(Getting 2 all four times)

P(Getting 2 exactly 3 times)

Now, lets find the three places out of 4 where we will get a 2. This can be done in 4C3 ways
=> \(\frac{4!}{3!*1!}\) = 4 ways

P(getting a 2) = \(\frac{1}{6}\) (as there is one way out of 6 in which we can get a 2)
P(getting any number except 2) = \(\frac{5}{6}\) (as we can get any of the 5 numbers, out of 6, except 2)

P(Getting 2 exactly 3 times) = P(Getting exactly three 2's and getting any number except 2 in 1 roll) = Number of ways * P(Getting 2) * P(Getting 2) * P(Getting 2) * P(Getting any number except 2) = 4 * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{5}{6}\) = 4*(\(\frac{1}{6}\))^3 * \(\frac{5}{6}\)

P(Getting 2 all four times)

P(Getting 2 all four times) = \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) = (\(\frac{1}{6}\))^4

=> P(getting a two at least 3 times) = P(Getting 2 exactly 3 times) + P(Getting 2 all four times) = 4*(\(\frac{1}{6}\))^3 * \(\frac{5}{6}\) + (\(\frac{1}{6}\))^4

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

“At least 3 twos in 4 rolls” means we could have 3 twos and 1 non-two number OR all 4 twos.

P(3 twos and 1 non-two number) = 4C3 x (1/6)^3 x (5/6) = 4(1/6)^3 (5/6)

P(all 4 twos) = 4C4 x (1/6)^4 = (1/6)^4

Therefore, P(at least 3 twos) = 4(1/6)^3 (5/6) + (1/6)^4

Answer: D