Question

Suppose X has a binomial distribution B$\bigg(6,\frac{1}{2}\bigg).$Show that X=3 is the most likely outcome.

Answer 1

B$\bigg(6, \frac{1}{2}\bigg)\Rightarrow$ n=6, p=$\frac{1}{2}$,q=$\frac{1}{2}$
xt=0,1,2,3,4,5,6

P(X=r)=C(n,r)$p^rq^{n-r}$

Since p=q

Therefore,P(X=r)=C(n,r)$p^n$

Now,out of C(6,0),C(6,1),C(6,2),C(6,3),C(6,4),C(6,5),C(6,6),here C(6,3)is maximum.

∴ P(X=3)=C(6,3)$\bigg(\frac{1}{2}\bigg)^6=\frac{6*5*4}{1*2*3}\bigg(\frac{1}{2}\bigg)^6=20*\frac{1}{64}=\frac{5}{16}$

Therefore,P(X=3)is maximum i.e., $\frac{5}{16}$