Question

Three Dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2 ?

• A. 36
• B. 81
• C. 91
• D. 116

Answer 1

Answer: (C)

91

There can be 3 cases :

I. When one dice shows $2$.

II. When two dice show $2$.

III. When three dices show $2.$

Case I: The dice which shows $2$ can be selected out of the $3$ dices in ${3_C}_1$ ways.

Remaining $2$ dices can have any $5$ numbers except $2$.

So number of ways for them = ${5_C}_1$ each, so no of ways when one dice shows $2=$ ${3_C}_1$ × ${5_C}_1$ × ${5_C}_1$.

Case II : Two dices, showing $2$ can be selected out of the $3$ dices in ${3_C}_2$ ways and the rest one can have any $5$ numbers except 2, so number of ways for the remaining $1$ dice $=5.$

So, number of ways, when two dices show $2=$ ${3_C}_2$ $\times 5$

Case III : When three dices show $2$ then these can be selected in ${3_C}_3$ ways.

So, number of ways, when three dices show $2=$${3_C}_3$ $=1$

As, either of these three cases are possible. Hence, total number of ways $=(3\times 5\times 5)+(3\times 5)+1=91$