MHT CET · Maths · Probability
If \(X \sim B(4, p)\) and \(P(X=0)=\frac{16}{81}\), then \(P(X=4)=\)
- A \(\frac{1}{81}\)
- B \(\frac{1}{16}\)
- C \(\frac{1}{8}\)
- D \(\frac{1}{27}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{81}\)
Step-by-step Solution
Detailed explanation
We have \(\mathrm{n}=4\) and \(\mathrm{P}(\mathrm{x}=0)=\frac{16}{81}\)
\(\begin{aligned}& \therefore \frac{16}{81}={ }^4 \mathrm{C}_0(\mathrm{p})^4(\mathrm{q})^0 \\& \therefore \frac{16}{81}=(\mathrm{p})^4 \Rightarrow \mathrm{P}=\frac{2}{3} \Rightarrow \mathrm{q}=\frac{1}{3} \\& \therefore \mathrm{P}(\mathrm{x}=4)={ }^4 \mathrm{C}_4(\mathrm{p})^0(\mathrm{q})^4=(1)(1)\left(\frac{1}{3}\right)^4=\frac{1}{81}\end{aligned}\)
\(\begin{aligned}& \therefore \frac{16}{81}={ }^4 \mathrm{C}_0(\mathrm{p})^4(\mathrm{q})^0 \\& \therefore \frac{16}{81}=(\mathrm{p})^4 \Rightarrow \mathrm{P}=\frac{2}{3} \Rightarrow \mathrm{q}=\frac{1}{3} \\& \therefore \mathrm{P}(\mathrm{x}=4)={ }^4 \mathrm{C}_4(\mathrm{p})^0(\mathrm{q})^4=(1)(1)\left(\frac{1}{3}\right)^4=\frac{1}{81}\end{aligned}\)
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