Question

Three six faced dice are tossed together, then the probability that exactly two of the three numbers are equal is

• A. $165/216$
• B. $177/216$
• C. $51/216$
• D. $90/216$

Answer 1

Answer: (D)

$90/216$

Let A, B and C be the three 6-faced dice.

Then, according to the question,

Since two dices has to be equal, that value can be any of the 6 faces, i.e., ${}^6{C}_1$ cases.

Now for each case, 2 equal dices can be selected from 3 dices in ${}^3{C}_2$ i.e., 3 ways.

And for each of the above, the third dice can have any of the 5 remaining faces

The possible outcomes are $P\left(A\right)=\frac{1}{6},P\left(B\right)=\frac{1}{6},P\left(C\right)=\frac{5}{6},P\left(A\right)=\frac{1}{6},P\left(B\right)=\frac{5}{6},P\left(C\right)=\frac{1}{6}$ and $P\left(A\right)=\frac{5}{6},P\left(B\right)=\frac{1}{6},P\left(C\right)=\frac{1}{6}$

Hence the required probability = $\frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times 6\times 3=\frac{90}{216}=\frac{5}{12}$