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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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