A die is thrown 2n + 1 times, `n in N` . The probability that faces with even numbers show odd number of times is :

Options

A. `(1)/(2)`
B. `lt(1)/(2)`
C. `gt(1)/(2)`
D. none of these

Answer

A

Explanation

We,have
p= Probability of getting an even number in a throw
`rArrp=(3)/(6)=(1)/(2)`
`therefore q=1-p=(1)/(2)`
Let X denote the number of times an even number is shown in `(2n+1)` throws of a dice. Then, X follows binomial distribution such that
`P(X=r)=.^2n+1C_(r)((1)/(2))^2n+1-r((1)/(2))^r=.^2n+1C_(r) ((1)/(2))^2n+1`.
`therefore` Required Probability `=sum_(r=1)^(2n+1)P(X=r)=sum _(r=1)^(2n+1)2n+1.^C_r((1)/(2))^2n+1`
`=((1)/(2))^2n+1{.^(2n+1)C_1+.^(2n+1)C_3+.....+.^(2n+1)C_(2n+1)}`
`=(1)/(2^(2n+1))xx2^2n=(1)/(2)`.