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Three numbers are chosen from 1 to 20. The probability that they are not consecutive isA. \(\frac{186}{190}\)B. \(\frac{187}{190}\) C. \(\frac{188}{190}\) D. \(\frac{18}{^{20}C_3}\) |
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Answer» 3 numbers can be chosen out of 20 in 20C3 ways. As we have to choose 3 consecutive numbers. So we can’t choose 19 and 18. Choosing consecutive number is same as choosing a single number because other 2 numbers are automatically chosen. This can be chosen 18 ways. ∴ P(chosen numbers are consecutive) = \(\frac{18}{{20}^C_3}\) Our answer matches with option (d) ∴ Option (d) is the only correct choice |
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