Question

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 tails is `((2n !))/((n !)^2)(1/2)^(2n)` b. `1-((2n !))/((n !)^2)` c. `1-((2n !))/((n !)^2)1/(4^n)^` d. none of these
A. `((2n!))/((n!)^(2))((1)/(2))^(2n)`
B. `1-((2n!))/((n!)^(2))`
C. `1-((2n!))/((n!)^(2))(1)/(4^(n))`
D. None of these

Answer 1

Correct Answer - C
The required probability is
1- probability of getting equal number of heads and tails
`=1-""^(2n)C_(n)((1)/(2))^(n)((1)/(2))^(2n-n)`
`=1-((2n)!)/((n!)^(2))xx(1)/(4^(n))`