Three dice are thrown together.Find the probability of getting a total of atleast 6
Solution
Correct Answer: Option B
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
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