Question

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

Answer 1

X is the random variable whose binomial distribution is B$(6,\frac{1}{2})$.

Therefore, n=6 and $p=\frac{1}{2}andq=1-p=1-\frac{1}{2}=\frac{1}{2}$

Then, P(X=r)= ${}^6{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{6-r}$

Here, P(X=0) = ${}^6{C}_0{p}^0{q}^6={}^6{C}_0{\left(\frac{1}{2}\right)}^6=\frac{1}{2^6}=\frac{6}{{64}^{\prime }}$

P(X=1) = ${}^6{C}_1{p}^1{q}^5=6\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^5=6\left(\frac{1}{2^6}\right)=\frac{6}{64}$

P(X=2) = ${}^6{C}_2{p}^2{q}^4=\frac{6\times 5}{1\times 2}{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^4=\frac{15}{{64}^{\prime }}$

P(X=3) = ${}^6{C}_3{p}^3{q}^3=\frac{6\times 5\times 4}{1\times 2\times 3}{\left(\frac{1}{2}\right)}^3={\left(\frac{1}{2}\right)}^3=\frac{20}{{64}^{\prime }}$

P(X=4) = ${}^6{C}_4{p}^4{q}^2={}^6{C}_2{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^2=\frac{15}{64}$

P(X=5) = ${}^6{C}_5{p}^5{q}^1={}^6{C}_1{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^1=\frac{6}{64}$

and P(X=6) = ${}^6{C}_6{p}^5{q}^0=1\times {\left(\frac{1}{2}\right)}^6=\frac{1}{64}$

It is clear that $\frac{20}{64}$ is maximum of all the above values. This means that X=3 is the most likely outcome.