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 )>1 / 2 is -

Step-by-step solution

Ways to make the sum K is coefficient of x^ K in (x+x² )¹⁰ Coefficient of x^ K in x¹⁰(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 aligned P ( K ) &= ¹⁰ C _ K -10 + ¹⁰ C _ K -9 + . .+ ¹⁰ C ₁₀ 2¹⁰ P ( K ) &= 2¹⁰- ( ¹⁰ C ₀+ + ¹⁰ C _ K -11 ) 2¹⁰ &=1- ¹⁰ C ₀+ . .+ ¹⁰ C _ K -11 2¹⁰ > 1 2 aligned 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 So K =16