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$\frac{1}{8}$
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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