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InterviewSolution
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Three distinct numbers are chosen at random from the first 15 natural numbers. The probability that the sum will be divisible by 3 isA. `(30)/(91)`B. `(31)/(91)`C. `(60)/(91)`D. none of these |
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Answer» Correct Answer - B Out of first 15 natural numbers, 3 natural numbers can be chosen is `.^(15)C_(3)` ways. When we divide a natural number by 3, it leaves either 0 or 1 or 2 as the remainder, so, let us divide 15 natural numbers into the following groups : `G_(1) : 1 4 7 10 13` `G_(2) : 2 5 8 11 14` `G_(3) : 3 6 9 12 15` The sum of three natural numbers chosen from these 15 natural numbers will be divisible by 3 if either all the three natural numbers are chosen from the same group or one natural is chosen from each group. `therefore` Favourable number of elementary events `=(.^(5)C_(3)+ .^(5)C_(3)+ .^(5)C_(3))+(.^(5)C_(1)xx.^(5)C_(1)xx .^(5)C_(1))=155` Hence, required probability `=(155)/(.^(15)C_(3))=(155xx6)/(15xx14xx13)=(31)/(91)` |
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