Question

For a binomial distribution : $n=6$ and $9$. $P(X=4)=P(X=2)$. Then we have

• A. $P(X=1)\le P(X=5)<P(X=3)$
• B. $P(X=5)>P(X=3)>P(X=1)$
• C. $P(X=5)<P(X=3)<P(X=1)$
• D. $P(X=5)+P(X=3)>P(X=1)$

Answer 1

The correct option is C $P(X=5)<P(X=3)<P(X=1)$

Given, $n=6$, $9P(X=4)=P(X=2)$.

which implies $9\times {{}^{6}C}_4\times p^4\times q^2={{}^{6}C}_2\times p^2\times q^4$, where $p$ is success and $q$ is failure. such that $p+q=1$.

therefore $9\times p^2=q^2,3\times p=q$

so we get $p=1/4,q=3/4$

Now $P(X=5)={{}^{6}C}_5\times {\left(1/4\right)}^5\times \left(3/4\right)=18/4^6$

$P(X=3)={{}^{6}C}_3\times {\left(1/4\right)}^3\times {\left(3/4\right)}^3=540/4^6$

$P(X=1)={{}^{6}C}_1\times {\left(1/4\right)}^1\times {\left(3/4\right)}^5=1458/4^6$

Therefore $P(X=5)<P(X=3)<P(X=1)$