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) = ^{n}C_{x}p^{x}q^{n-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 ^{6}C_{x} is maximum.

Let us check at what value of x would ^{6}C_{x} be maximum.

Check at x = 0.

⇒ ^{6}C_{0} = 1

Check at x = 6.

⇒ ^{6}C_{6} = 1

Check at x = 1.

⇒ ^{6}C_{1} = 6

Check at x = 5.

⇒ ^{6}C_{5} = 6

Check at x = 2.

⇒ ^{6}C_{2} = 15

Check at x = 4.

⇒ ^{6}C_{4} = 15

Check at x = 3.

⇒ ^{6}C_{3} = 20

Note that at x = 3, ^{6}C_{x} is maximum.

Hence, we have shown that x = 3 is the most likely outcome.

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