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Question: A coin is tossed 2n times. The chance that the number of times one gets head is not equal to the num...

A coin is tossed 2n times. The chance that the number of times one gets head is not equal to the number of times one gets tail is

A

(2n!)(n!)2(12)2n\frac { ( 2 n ! ) } { ( n ! ) ^ { 2 } } \left( \frac { 1 } { 2 } \right) ^ { 2 n }

B

1−(2n!)(n!)21 - \frac { ( 2 n ! ) } { ( n ! ) ^ { 2 } }

C

1−(2n!)(n!)2⋅14n1 - \frac { ( 2 n ! ) } { ( n ! ) ^ { 2 } } \cdot \frac { 1 } { 4 ^ { n } }

D

None of these

Answer

1−(2n!)(n!)2⋅14n1 - \frac { ( 2 n ! ) } { ( n ! ) ^ { 2 } } \cdot \frac { 1 } { 4 ^ { n } }

Explanation

Solution

The required probability

= 1 – Probability of equal number of heads and tails

=1−2nCn(12)n(12)2n−n=1−(2n)!n!n!(14)n=1−(2n)!(n!)2⋅14n= 1 - { } ^ { 2 n } C _ { n } \left( \frac { 1 } { 2 } \right) ^ { n } \left( \frac { 1 } { 2 } \right) ^ { 2 n - n } = 1 - \frac { ( 2 n ) ! } { n ! n ! } \left( \frac { 1 } { 4 } \right) ^ { n } = 1 - \frac { ( 2 n ) ! } { ( n ! ) ^ { 2 } } \cdot \frac { 1 } { 4 ^ { n } } .