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Question

Suppose X has a binomial distribution B of 6,12. 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)

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Solution

X is the random variable whose binomial distribution is B(6,12).
Therefore, n=6 and p=12
q=1p=112
=12
Then, P(X=x)=nCxqnxpx
=6Cx(12)6x(12)x
=6Cx(12)6
It can be seen that P(X=x) will be maximum, if 6Cx will be maximum.
Then, 6C0=6C0=6!0!6!=1
6C1=6C5=6!1!5!=6
6c2=6C4=6!2!4!=15
6C3=6!3!3!=20
The value of 6C3 is maximum. Therefore, for x=3,P(X=3) is maximum.
Thus, X=3 is the most likely outcome.

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