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Question

Two dice were thrown simultaneously. Then find the probability of getting at least one outcome as multiple of 3.

A
29
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B
16
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C
512
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D
59
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Solution

The correct option is D 59
The total possible outcomes will be 6×6=36
They are :
(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),
(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),
(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),
(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),
(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),
(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)
Number of getting atleast one outcome as multiple of 3=20
[i.e.(1,3)(1,6)(2,3)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,6)(5,3)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)]
P(A)=Number of favorable outcomesTotal Number of possible outcomes
Probability P(A)=2036=59
hence, option D is correct.

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