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Let `omega`be a complex cube root unity with `omega!=1.`A fair die is thrown three times. If `r_1, r_2a n dr_3`are the numbers obtained on the die, then the probability that `omega^(r1)+omega^(r2)+omega^(r3)=0`is`1//18`b. `1//9`c. `2//9`d. `1//36`A. `(1)/(18)`B. `(1)/(9)`C. `(2)/(9)`D. `(1)/(36)` |
Answer» Correct Answer - C Clearly, Total number of elementary events `=6xx6xx6=216` Clearly, `w^(r_(1))+w^(r_(2))+w^(r_(3))=0`, if one of `r_(1), r_(2) " and " r_(3)` takes values from the set {3,6}, other takes values from the set {1,4} and the third takes values from the set {2,5}. The total number of these ways is `(.^(2)C_(1)xx .^(2)C_(1)xx .^(2)C_(1))xx3!` So, favourable number of elementary events `={.^(2)C_(1)xx .^(2)C_(1)xx .^(2)C_(1))xx3!=48` Hence, required probability `=(48)/(216)=(2)/(9)` |
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