Question

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 ( x i ), x i = 0, 1, 2, 3, 4, 5, 6)

Answer 1

It is given that X has a binomial distribution B( 6, 1 2 ).

It can be observed that the Bernoulli trials is n=6 and probability in each trial is p= 1 2 .

The property of probability gives that p+q=1.

So,

q=1−p =1− 1 2 = 1 2

The probability of x successes P( X=x ) is,

P( X=x )= C n x q n−x p x = C 6 x ( 1 2 ) 6−x ( 1 2 ) x = C 6 x ( 1 2 ) 6

Where x=0,1,…,n

The maximum value of P( X=x )is obtained by the maximum value of C 6 x .

So,

C 6 0 = C 6 6 = 6! 0!6! =1

C 6 1 = 6! 1!5! =6

C 6 2 = 6! 2!4! =15

C 6 3 = 6! 3!3! =20

The maximum value is 20 at C 6 3 .

So, P( X=x ) is maximum. for x=3.

Therefore, the most likely outcome is X=3.