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Question: A coin is thrown 2n+1 times. Probability of getting a prime number atmost n times is ..... P = \(\f...

A coin is thrown 2n+1 times. Probability of getting a prime number atmost n times is ..... P = 12,q=12\frac { 1 } { 2 } , \mathrm { q } = \frac { 1 } { 2 }

A

1/n

B

1/2n

C

½

D

n/n+1

Answer

½

Explanation

Solution

P(X≤ n)

= ∑r=0n2n+1Cr122n+1=2n+1C0+2n+1C1+2n+1C2+……..2n+1Cn22n+1\sum _ { r = 0 } ^ { n } { } ^ { 2 n + 1 } C _ { r } \frac { 1 } { 2 ^ { 2 n + 1 } } = \frac { { } ^ { 2 n + 1 } C _ { 0 } + { } ^ { 2 n + 1 } C _ { 1 } + { } ^ { 2 n + 1 } C _ { 2 } + \ldots \ldots . . { } ^ { 2 n + 1 } C _ { n } } { 2 ^ { 2 n + 1 } }

= 12⋅22n+122n+1=12\frac { \frac { 1 } { 2 } \cdot 2 ^ { 2 n + 1 } } { 2 ^ { 2 n + 1 } } = \frac { 1 } { 2 }