InterviewSolution
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InterviewSolution
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Three dice are thrown together. The probability that the sum of the numbers appearing on them is 9, isA. `(7)/(216)`B. `(47)/(54)`C. `(7)/(54)`D. `(7)/(9)` |
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Answer» Correct Answer - C Total number of elementary events `=6^(3)`. Total number of ways of getting 9 as the sum = Coefficient of `x^(9) " in " (x^(1)+x^(2)+x^(3)+x^(4)+x^(5)+x^(6))^(3)` = Coefficient of `x^(6) " in " (1+x+x^(2)+x^(3)+x^(4)+x^(5))^(3)` = Coefficient of `x^(6) " in " ((1=x^(6))/(1-x))^(3)` = Coefficient of `x^(6) " in " (1-x)^(-3)= .^(6+3-1)C_(3-1)= .^(8)C_(2)=28` `therefore` Favourable number of elementary events =28. So, required probability `=(28)/(216)=(7)/(54)` |
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