Question

A fair coin is tossed 100 times. The probability of getting tails 1, 3, .., 49 times is `1//2` b. `1//4` c. `1//8` d. `1//16`
A. `1//2`
B. `1//4`
C. `1//8`
D. `1//16`

Answer 1

Correct Answer - B
Let the probability of getting a tail in a single trial be `p=1//2.` The number of trials be n=100 and the number of trials in 100 trials be X.
`P(X=r)=""^(100)C_(r)p^(r)q^(n-r)`
`=""^(100)C_(r)((1)/(2))^(r)((1)/(2))^(100-r)=""^(100)C_(r)((1)/(2))^(100)`
Now, `N P(X=1)+P(X=3)+...+P(X=49)`
`=""^(100)C_(1)((1)/(2))^(100)+""^(100)C_(3)((1)/(2))^(100)+...+""^(100)C_(49)((1)/(2))^(100)`
`But ""^(100)C_(1)+""^(100)C_(3)+...+""^(100)C_(99)=2^(99)`
Also, `""^(100)C_(99)=""^(100)C_(1)`
`""^(100)C_(97)=""^(100)C_(3)...""^(100)C_(51)=""^(100)C_(49)`
Thus, `2(""^(100)C_(1)+""^(100)C_(3)+...+""^(100)C_(49)=2^(99)`
`or""^(100)C_(1)+""^(100)C_(3)+...+""^(100)C_(49)=2^(98)`
Therefore, probability of required event is
`(2^(98))/(2^(100))=(1)/(4)2^(98)//2^(100)=1//4`