Q83 / 125 · BITSAT 2015 · Mathematics

If in a binomial distribution $\mathrm{n}=4, \mathrm{P}(\mathrm{X}=0)=\frac{16}{81}$, then $\mathrm{P}(\mathrm{X}=4)$ equals

  • A
    $\frac{1}{16}$
  • Correct answer
    B
    $\frac{1}{81}$
  • C
    $\frac{1}{27}$
  • D
    $\frac{1}{8}$

Solution

Given $n=4$ and $P(X=0)=\frac{16}{81}$ Let $\mathrm{p}$ be the probability of success and $\mathrm{q}$ that of failure in a trial. $$ \text { Then, } \mathrm{P}(\mathrm{X}=0)={ }^{4} \mathrm{C}_{0} \mathrm{p}^{0} \mathrm{q}^{4}=\frac{16}{81} $$ $\Rightarrow \quad \mathrm{q}^{4}=\left(\frac{2}{3}\right)^{4}$ $\left(\because{ }^{n} C_{0}=1\right)$ $\Rightarrow \quad \mathrm{q}=\frac{2}{3} \Rightarrow \mathrm{p}=\frac{1}{3}$ $\therefore \quad \mathrm{P}(\mathrm{X}=4)={ }^{4} \mathrm{C}_{4} \mathrm{p}^{4} \mathrm{q}^{0}=\mathrm{p}^{4}=\left(\frac{1}{3}\right)^{4}=\frac{1}{81}$

Hence, the correct answer is option: 'B'.

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