Suppose X has a binomial distribution . Show that X = 3 is the most likely outcome.

(Hint: P(X = 3) is the maximum among all P(xi), xi = 0,1,2,3,4,5,6)

As per the question,

X is any random variable whose binomial distribution is

Thus, P(X = x) = ^{n}C_{x}q^{n-x}p^{x} , where x = 0, 1, 2, …n

It can be clearly observed that P(X = x) will be maximum if ^{6}c_{x} will bw maximum.

∴^{6}c_{x} = ^{6}c_{6} = 1

^{6}c_{1} = ^{6}c_{5} = 6

^{6}c_{2} = ^{6}c_{4} = 15

^{6}c_{3} = 20

Hence we can clearly see that ^{6}c_{3} is maximum.

∴for x = 3, P(X = x) is maximum.

Hence, proved that the most likely outcome is x = 3.

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