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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) 112
(E) 116

We need to determine, when rolling three dice, the number of outcomes in which at least one die shows a 2.

We should first recall the following formula:

Total number of ways to roll 3 dice = number of ways with at least one 2 + number of ways with no 2s.

Since there are 6 x 6 x 6 = 216 possible outcomes when rolling 3 dice, we have:

216 = number of ways with at least one 2 + number of ways with no 2s

number of ways with at least one 2 = 216 - number of ways with no 2s

There are 5 x 5 x 5 = 125 outcomes in which no 2 is shown.

Thus, there are 216 - 125 = 91 outcomes in which at least one 2 is shown.

Answer: C
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For 1 time 2 -

1*5*5*3 = 75

For 2 times = 1*1*5*3 = 15

For 3 times = 1 = 75+15+1 = 91.
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saswata4s
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) 112
(E) 116

Hi

#1 Direct counting:

one 2 + two 2's + three 2's = 1*5*5*3!/2! + 1*1*5*3!/2! + 1*1*1 = 75 + 15 + 1 = 91

# Indirect counting:

Total # of outcomes - # of outcomes without 2's = \(6^3 - 5^3 = 91\)

Answer C.

Posted from my mobile device

Why do we divide by 2! In the direct counting method?
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saswata4s
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) 112
(E) 116

total possible cases of getting atleast 2 in 3 dice rolls
1 is depicting value 2 and 5 is for values of other places except 2

(1,5,5) ; 25*3 ; 75
(1,1,5) ; 5*3 ; 15
(1,1,1) ; 1
total 91
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