# For a binomial distribution mean is 6 and variance is

For a binomial distribution mean is 6 and variance is 2. Find the $$\displaystyle{P}{\left({X}={4}\right)}$$.
a) 0.1024
b) 0.0012
c) 0.1768
d) 0.1345
e) 0.8976

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Arnold Odonnell
Step 1
Given :
For a binomial distribution mean is 6 and variance is 2.
Also,
The mean of the distribution $$\displaystyle{\left(\mu{x}\right)}$$ is equal to $$\displaystyle{n}\cdot{P}={6}$$
The variance $$\displaystyle{\left({\sigma_{{{x}}}^{{{2}}}}\right)}\ {i}{s}\ {n}\cdot{p}\cdot{\left({1}-{p}\right)}={2}$$
Step 2
We can put the value of $$\displaystyle{n}\cdot{p}$$ in variance equation ,then
$$\displaystyle{n}\cdot{p}\cdot{\left({1}-{p}\right)}={2}$$
$$\displaystyle{6}\cdot{\left({1}-{p}\right)}={2}$$
$$\displaystyle{1}-{p}=\frac{{1}}{{3}}$$
$$\displaystyle{p}=\frac{{2}}{{3}}$$
Hence,
$$\displaystyle{n}=\frac{{6}}{{\frac{{2}}{{3}}}}={9}$$
The full binomial probability formula with the binomial coefficient is:
$$\displaystyle{P}{\left({X}\right)}={\frac{{{n}!}}{{{X}!{\left({n}-{X}\right)}!}}}\cdot{p}^{{{X}}}\cdot{\left({1}-{p}\right)}^{{{n}-{X}}}$$
Computing $$\displaystyle{P}{\left({X}={4}\right)}$$
$$\displaystyle{P}{\left({4}\right)}={\frac{{{9}!}}{{{4}!{\left({9}-{4}\right)}!}}}\cdot{0.6667}^{{{4}}}\cdot{\left({1}-{0.6667}\right)}^{{{9}-{4}}}$$
$$\displaystyle{P}{\left({X}={4}\right)}={0.10239}\approx{0.1024}$$
Correct Option is:
a) 0.1024.