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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 `r_(1),r_(2), r_(3) in {1, 2, 3, 4, 5, 6}` `r_(1), r_(2), r_(3)` are of the form 3k, 3k + 1, 3k + 2 Required probability = `(3! xx .^(2)C_(1)xx.^(2)C_(1) xx.^(2)C_(1))/(6xx6xx6) = (6xx8)/(216) = (2)/(9)` |
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