InterviewSolution
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InterviewSolution
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A man tosses a coin 10 times, scoring 1 point for each head and 2 points for each tail. Let P(K) be the probability of scoring at least K points. The largest value of K such that `P(K)gt1//2` is -A. 14B. 15C. 16D. 17 |
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Answer» Correct Answer - C Ways to make the sum K is coefficient of `x^(K)` in `(x+x^(2))^(10)` Coeffecient of `x^(K)` in `x^(10)(1+x)^(10)` Coefficient of `x^(K-10)` in `(1+x)^(10)` Which is `.^(10)C_(K-10)` So ways to make sum minimum K is `.^(10)C_(K-10)+.^(10)C_(K-9)+.^(10)C_(K-8)+......^(10)C_(10)` Probability `P(K)=(.^(10)C_(K-10 )+.^(10)C_(K-9)+.......+.^(10)C_(10))/(2^(10))` `P(K)=(2^(10)-(.^(10)C_(0)+.....+.^(10)C_(K-11)))/(2^(10))` `=I-(.^(10)C_(0)+....+.^(10)C_(K-11))/(2^(10))gt(1)/(2)` But K should be maximum so `.^(10)C_(K-11)=.^(10)C_(5)` (middle value) So that `.^(10)C_(0)+.....+.^(10)C_(K-11)` is max So K = 16 |
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