Three dice (each having six faces with each face having one number

Number of possible outcomes such that at least one dice shows the number 2 = Total Outcomes - All outcomes without any 2

Total Outcomes on three dice = 6*6*6 = 216

Total Outcomes Without any 2 = 5*5*5 (as all 5 outcomes except "2" on each die is acceptable) = 125

Required Outcomes = 216 - 125 = 91

Answer: Option C
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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.

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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Why do we divide by 2! In the direct counting method?

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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