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Given that Two fair dice are thrown together and we need to find what is the probability that the number obtained on one of the dice is a multiple of the number obtained on the other die?

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

Lets start writing the possible cases where number of one dice is a multiple of other

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) [ All cases with 1 in the first roll are possible as all numbers are multiples of 1 ]
(2,1), (2,2), (2,4), (2,6)
(3,1), (3,3), (3,6)
(4,1), (4,2), (4,4)
(5,1), (5,5)
(6,1), (6,2), (6,3), (6,6)

=> 22 cases

=> Probability that the number obtained on one of the dice is a multiple of the number obtained on the other die = \(\frac{22}{36}\) = \(\frac{11}{18}\)

So, Answer will be A
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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rverma007
Two fair dice are thrown together.
­When 2 fair dice thrown together, sample space is 36.
Quote:
the number obtained on one of the dice is a multiple of the number obtained on the other die.
For such 2 numbers (_ _) only possible combinations are as below
for 1 - there are total 11 posibilities (1,1) (1,2) ..... (1,6) (6,1) (5,1)....... (2,1)
for 2, there are 5 possibilities - (2,2) (2,4) (2,6) (4,2) (6,2)
similarly for 3 there will be 3 (3,3) (3,6) (6,3)
and for 4,5,6 there will be 1 for each so total - 3
So total possibilities = 11 + 5 + 3 + 3 = 22
Probability = 22/36 = 11/18
Answer is A.
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