In a binomial distribution `B(6,1/2)`,
`n = 6 and p = 1/2`
Here, `n` is number of the trials and `p` is the success probability of each trial.
`:. q = 1- 1/2 = 1/2.`
Now, `P(X = k) = C(6,k) (1/2)^k(1/2)^(6-k) = (1/2)^6 C(6,k)`
So, most likely outcome will have maximum value of `C(6,k)`.
When `k = 0, C(6,0) = 1`
When `k = 1, C(6,1) = 6`
When `k = 2, C(6,2) = (6**5)/(2**1) = 15`
When `k = 3, C(6,3) = (6**5**4)/(3**2**1) = 20`
When `k = 4, C(6,4) = (6**5)/(2**1) = 15`
When `k = 5, C(6,5) = 6`
When `k = 6, C(6,6) = 1`
`:. C(6,3)` is maximum.
So, most likely outcome is `X=3`.