Suppose X has a binomial distribution `B(6,""""1/2)` . Show that `X" "=" "3` is the most likely outcome. (Hint: `P(x=3)` is the maximum among all `P(x_i),""x_i=0,""1,""2,""3,""4,""5,""6)`

Answer

N/a

Explanation

Let `X` be a random variable whose binomial distribution is `B(6,1/2)`
Therefore `n=6` and `p=1/2`
and `q=1-p=1-1/2=1/2`
Then `P(X=r)=.^(6)C_(r).(1/2)^(r)(1/2)^(6-r)`
Here
`P(X=0)=.^(6)C_(0)p^(0)q^(6)=.^(6)C_(0)(1/2)^(6)=1/(2^(6))=1/64`
`P(X=1)=.^(6)C_(1)p^(1)q^(5)=6(1/2)(1/2)^(5)=6(1/(2^(6)))=6/64`
`P(X=2)=.^(6)C_(2)p^(2)q^(4)=(6xx5)/(1xx2).(1/2)^(2)(1/2)^(4)=15/64`
`P(X=3)=.^(6)C_(3)p^(3)q^(3)=(6xx5xx4)/(1xx2xx3)(1/2)^(3)(1/2)^(3)=20/64`
`P(X=4)=.^(6)C_(4)p^(4)q^(2)=.^(6)C_(2)(1/2)^(4)(1/2)^(2)=15/64`
`P(X=5)=.^(6)C_(5)p^(4)q^(1)=.^(6)C_(1).(1/2)^(5)(1/2)^(1)=6/64`
and `P(X=6)=.^(6)C_(6)p^(6)q^(0)=1xx(1/2)^(6)=1/64`
Clearly, the maximum value among all results is `20/64`
Therefore, `X=3` is the result with maximum probability.