Q32 / 90 · KVPY 2012 (SB/SX) · Mathematics
A man tosses a coin 10 times, scoring 1 point for each head and 2 points for each tail. Let $\mathrm{P}(\mathrm{K})$ be the probability of scoring at least $\mathrm{K}$ points. The largest value of $\mathrm{K}$ such that $\mathrm{P}(\mathrm{K})>1 / 2$ is $-$
- A14
- B15
- Correct answerC16
- D17
Solution
Ways to make the sum $K$ is coefficient of $x^{K}$ in $\left(x+x^{2}\right)^{10}$
Coefficient of $x^{K}$ in $x^{10}(1+x)^{10}$
Coefficient of $x^{K-10}$ in $(1+x)^{10}$
Which is ${ }^{10} \mathrm{C}_{\mathrm{K}-10}$
So ways to make sum minimum $\mathrm{K}$ is
${ }^{10} \mathrm{C}_{\mathrm{K}-10}+{ }^{10} \mathrm{C}_{\mathrm{K}-9}+{ }^{10} \mathrm{C}_{\mathrm{K}-8}+\ldots . .+{ }^{10} \mathrm{C}_{10}$
Probability
$\begin{aligned}
\mathrm{P}(\mathrm{K}) &=\frac{{ }^{10} \mathrm{C}_{\mathrm{K}-10}+{ }^{10} \mathrm{C}_{\mathrm{K}-9}+\ldots \ldots \ldots \ldots . .+{ }^{10} \mathrm{C}_{10}}{2^{10}} \\
\mathrm{P}(\mathrm{K}) &=\frac{2^{10}-\left({ }^{10} \mathrm{C}_{0}+\ldots \ldots+{ }^{10} \mathrm{C}_{\mathrm{K}-11}\right)}{2^{10}} \\
&=1-\frac{{ }^{10} \mathrm{C}_{0}+\ldots \ldots . .+{ }^{10} \mathrm{C}_{\mathrm{K}-11}}{2^{10}}>\frac{1}{2}
\end{aligned}$
But $\mathrm{K}$ should be maximum so ${ }^{10} \mathrm{C}_{\mathrm{K}-11}={ }^{10} \mathrm{C}_{5}$ (middle value) So that ${ }^{10} \mathrm{C}_{0}+\ldots .+{ }^{10} \mathrm{C}_{\mathrm{K}-11}$ is $\max$ So $\mathrm{K}=16$
Hence, the correct answer is option: 'C'.