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(1/6)*(5/6)*2! = 5/18
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What is the probability of rolling two normal six-sided dice and getting exactly one six?

A) 1/36
B) 1/6
C) 5/18
D) 11/36
E) 1/3

Number of outcomes/ Dice = 6, Total outcomes = 6^2 = 36

For getting exactly one SIX, the possibilities are like below -

(1,6)
(2,6)
.
.
(5,6)
(6,1)
(6,2)
.
.
(6,5) ............... total 10

Thus P(A) = 10/36 = 5/18 (Answer-C)
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We need to find What is the probability of rolling two normal six-sided dice and getting exactly one six?

As we are rolling two dice => Number of cases = \(6^2\) = 36

Now to get exactly one 6 we can get it in the first roll or the second roll.

Getting 6 in the first roll and not getting two sixes we will have 5 cases
(6,1), (6,2), (6,3), (6,4), (6,5)

Getting 6 in the second roll and not getting two sixes we will have 5 cases
(1,6), (2,6), (3,6), (4,6), (5,6)

=> 10 cases

=> Probability of getting exactly one six = \(\frac{10}{36}\) = \(\frac{5}{18}\)

So, Answer will be C
Hope it helps!

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