A fair coin is tossed 100 times. The probability of getting tails an odd number of times is `1//2` b. `1//8` c. `3//8` d. none of these

Options

A. `(1)/(2)`
B. `(3)/(8)`
C. `(1)/(4)`
D. `(1)/(8)`

Answer

A

Explanation

Let x denote number of tails . Then , X is a binomial varaite with parameters:
`x = 100` & ` p = (1)/(2)`
` therefore P(x =r) = ""^(100)C_(r)((1)/(2))^(100), ( r= 0, 1,2,……100). `
Req. probability `= P(x =1) + P(x =3) + …. +P(x = 99) `=[(1)/(2)]^(100) ["^(100)C_(1) +"^(100)C_(3)+....+"^(100)C_(99)]`
`= (1)/(2^(100))2^(99)=(1)/(2)`