Sawaal
17
Q:

Three dice are thrown together.Find the probability of getting a total of atleast 6.

A) 103/216 B) 103/108
C) 103/36 D) 36/103

Answer:   B) 103/108



Explanation:

Total number of events=6 x 6 x 6=216
Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.

So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)}

Favourable number of cases=10

Therefore,P(B)=10/216

=> P(A) = 1 - P(B)

             = 1 - (10/216) = 103/108

Subject: Probability
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