KVPY SX SB Exam 2012 Question Paper

Question : 89 of 120

Marks: +1, -0

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

Validate

Solution:

Ways to make the sum K is coefficient of

x^K

(x + x^2)^10

Coefficient of

x^K

x^10(1+ x)^10

Coefficient of

x^{K-10}

(1 +x)^10

Which is

^10C_{K-10}

So ways to make sum minimum K is

^10C_{K-10}+^10C_{K-9}

+^10C_{K-8}+...+^10C_10

Probability

P(K)={^10C_{K-10}+^10C_{K-9} +^10C_10}/2^10

P(K)={2^10-(^10C_0+....+^10C_{K-11})}/2^10

=1-{^10C_0+....+^10C_{K-11}}/2^10> 1/2

But K should be maximum so

^10C_{K-11}=^10C_5

(middle value)
So that

^10C_0+ ....+^10 C_{K- 11}

is max
So K = 16

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