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
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(c) There can be 3 cases :
I. When one dice shows 2.
II. When two dice shows 2.
III. When three dices shows 2.
Case I : The dice which shows 2 can be selected out of the 3 dices in 3C1 ways.
Remaining 2 dices can have any 5 numbers except 2. So number of ways for them = 5C1 each, so no of ways when one dice shows 2 = 3C1 × 5C1 × 5C1.
Case II : Two dices, showing 2 can be selected out of the 3 dices in 3C2 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 = 3C2 × 5
Case III : When three dices show 2 then these can be selected in 3C3 ways.
So, number of ways, when three dices show 2 = 3C3 = 1
As, either of these three cases are possible.
Hence, total number of ways
= (3 × 5 × 5) + (3 × 5) + 1 = 91
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