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Suppose X has a binomial distribution with n=6 and p=1/2. Show that X = 3 is the most likely outcome.
Given that, X has a binomial distribution.
Also, n = 6 & p = 1/2
And as we know that, p + q = 1
⇒ q = 1 – p
Let us define a binomial distribution with x as a variable out of other n variables.
P (X = x) = nCxpxqn-x
Putting values of n, p and q, we get
From the derived result, it can be seen that P (X = x) will be maximum if 6Cx is maximum.
Let us check at what value of x would 6Cx be maximum.
Check at x = 0.
⇒ 6C0 = 1
Check at x = 6.
⇒ 6C6 = 1
Check at x = 1.
⇒ 6C1 = 6
Check at x = 5.
⇒ 6C5 = 6
Check at x = 2.
⇒ 6C2 = 15
Check at x = 4.
⇒ 6C4 = 15
Check at x = 3.
⇒ 6C3 = 20
Note that at x = 3, 6Cx is maximum.
Hence, we have shown that x = 3 is the most likely outcome.