MCQ
If in a binomial distribution $\text{n}=4,\text{P(X}=0)=\frac{16}{81},$ then $\text{P(X}=4)$ equals:
  • A
    $\frac{1}{16}$
  • B
    $\frac{1}{81}$
  • C
    $\frac{1}{27}$
  • D
    $\frac{1}{8}$

Answer

  1. $\frac{1}{81}$

Solution:

Given $\text{n}=4,\text{P(X}=0)=\frac{16}{81}$

$\text{P(X}=0)=\frac{16}{81}$

$\text{ }^5\text{C}_0\text{p}^0\text{q}^4=\frac{16}{81}$

$\text{q}^4=\frac{16}{81}$

$\text{q}=\frac{2}{3}\Rightarrow\text{p}=\frac{1}{3}$

$\Rightarrow\text{P(X}=4)=\text{ }^5\text{C}_4\big(\frac{1}{4}\big)^4=\frac{1}{81}$

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