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A special dice with numbers 1,-1,2,-2,0 and 3 is thrown thrice. What is the probability that the sum of the numbers occurring on the upper face is zero?

Answer» Below are the cases, when sum of numbers of all dice can be `0`.
`(i) (0,0,0) - 1` possibility
`(ii)(1,0,-1) - 3! = 6` posiibilities
`(iii)(2,0,-2) - 3! = 6` posiibilities
`(iv)(1,1,-2) - (3!)/2= 3` possibilities
`(v)(-1,-2,3) - 3! = 6` posiibilities
`(vi)(2, -1,-1) -(3!)/2= 3` possibilities
So, total number of favourable outcomes, `n(E) = 1+6+6+3+6+3 = 25`
Total number of outcomes `n(S) = 6^3 = 216`
`:.` Required probability, `P(E) = (n(E))/(n(S)) = 25/216`


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