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 -

Options

A. 14
B. 15
C. 16
D. 17

Answer

C

Explanation

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