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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Probability

BITSAT 2015 · Mathematics · Probability

If in a binomial distribution n=4, P(X=0)=(16)/(81) , then P(X=4)...

If in a binomial distribution n=4,P(X=0)=1681\mathrm{n}=4, \mathrm{P}(\mathrm{X}=0)=\frac{16}{81}, then P(X=4)\mathrm{P}(\mathrm{X}=4) equals

Options

  1. A116\frac{1}{16}
  2. B181\frac{1}{81}
  3. C127\frac{1}{27}
  4. D18\frac{1}{8}

Correct answer

B. 181\frac{1}{81}

Step-by-step solution

Given n=4n=4 and P(X=0)=1681P(X=0)=\frac{16}{81} Let p\mathrm{p} be the probability of success and q\mathrm{q} that of failure in a trial.  Then, P(X=0)=4C0p0q4=1681⇒q4=(23)4(∵(n0)=1)⇒q=23⇒p=13∴P(X=4)=4C4p4q0=p4=(13)4=181\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\binom{n}{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}
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