MHT CET · Maths · Probability
If \(X \sim B(4, p)\) and \(2 P(X=3)=3 P(X=2)\) then value of \(p\) is
- A \(\frac{9}{13}\)
- B \(\frac{4}{13}\)
- C \(\frac{1}{13}\)
- D \(\frac{12}{13}\)
Answer & Solution
Correct Answer
(A) \(\frac{9}{13}\)
Step-by-step Solution
Detailed explanation
Given \(X \sim B(4, p)\) and \(2 P(X=3)=3 P(X=2)\)
\(2\left[{ }^{4} C_{3} p^{3} q^{1}\right]=3\left[{ }^{4} C_{2} p^{2} q^{2}\right]\)
\(2 \times 4 p^{3} q=3 \times 6 p^{2} q^{2}\)
\(\therefore p=\frac{9 q}{4} \Rightarrow p=\frac{9}{4}(1-p) \Rightarrow p=\frac{9}{4}-\frac{9}{4} p\)
\(\left(1+\frac{9}{4}\right) p=\frac{9}{4} \Rightarrow \frac{13 p}{4}=\frac{9}{4} \Rightarrow p=\frac{9}{13}\)
\(2\left[{ }^{4} C_{3} p^{3} q^{1}\right]=3\left[{ }^{4} C_{2} p^{2} q^{2}\right]\)
\(2 \times 4 p^{3} q=3 \times 6 p^{2} q^{2}\)
\(\therefore p=\frac{9 q}{4} \Rightarrow p=\frac{9}{4}(1-p) \Rightarrow p=\frac{9}{4}-\frac{9}{4} p\)
\(\left(1+\frac{9}{4}\right) p=\frac{9}{4} \Rightarrow \frac{13 p}{4}=\frac{9}{4} \Rightarrow p=\frac{9}{13}\)
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