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)`

Suppose X has a binomial distribution `B(6,1/2)`. Show that `X = 3`is

1.	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» 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) = (65)/(21) = 15` When `k = 3, C(6,3) = (654)/(321) = 20` When `k = 4, C(6,4) = (65)/(21) = 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`.

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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)`

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