KCET · Maths · Probability
If a random variable \(X\) follows the binomial distribution with parameters \(n=5, p\) and \(P(X=2)=9 P(X=3)\), then \(p\) is equal to
- A \(10\)
- B \(1 / 10\)
- C \(5\)
- D \(1 / 5\)
Answer & Solution
Correct Answer
(B) \(1 / 10\)
Step-by-step Solution
Detailed explanation
\(\because p(X=2)=9 \times p(X=3)\) (where, \(n=5\) and \(q=1-p)\)
\(\Rightarrow \quad{ }^5 C_2 p^2(1-p)^3=9 \cdot{ }^5 C_3 p^3(1-p)^2\)
\(\Rightarrow \frac{5!}{2!3!} p^2(1-p)^3=9 \cdot \frac{5!}{3!2!} p^3(1-p)^2\)
\(\Rightarrow \quad \frac{p^2(1-p)^3}{p^3(1-p)^2}=9 \Rightarrow \frac{1-p}{p}=9\)
\(\Rightarrow \quad 9 p+p=1 \quad \Rightarrow \quad 10 p=1\)
\(\therefore \quad p=\frac{1}{10}\)
\(\Rightarrow \quad{ }^5 C_2 p^2(1-p)^3=9 \cdot{ }^5 C_3 p^3(1-p)^2\)
\(\Rightarrow \frac{5!}{2!3!} p^2(1-p)^3=9 \cdot \frac{5!}{3!2!} p^3(1-p)^2\)
\(\Rightarrow \quad \frac{p^2(1-p)^3}{p^3(1-p)^2}=9 \Rightarrow \frac{1-p}{p}=9\)
\(\Rightarrow \quad 9 p+p=1 \quad \Rightarrow \quad 10 p=1\)
\(\therefore \quad p=\frac{1}{10}\)
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