Question:

If a random variable XX follows the binomial distribution with parameters n=5n = 5, pp, and P(X=2)=9P(X=3)P(X = 2) = 9P(X = 3), then pp is equal to:

Updated On: Mar 29, 2025
  • 1010
  • 110\frac{1}{10}
  • 55
  • 1151\frac{1}{5}
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The Correct Option is B

Approach Solution - 1

1. Understand the binomial distribution:

Given XBinomial(n=5,p) X \sim \text{Binomial}(n=5, p) and P(X=2)=9P(X=3) P(X=2) = 9P(X=3) .

2. Write the binomial probabilities:

P(X=2)=(52)p2(1p)3=10p2(1p)3 P(X=2) = \binom{5}{2} p^2 (1-p)^3 = 10 p^2 (1-p)^3 P(X=3)=(53)p3(1p)2=10p3(1p)2 P(X=3) = \binom{5}{3} p^3 (1-p)^2 = 10 p^3 (1-p)^2

3. Set up the equation:

Given P(X=2)=9P(X=3) P(X=2) = 9P(X=3) :

10p2(1p)3=910p3(1p)2 10 p^2 (1-p)^3 = 9 \cdot 10 p^3 (1-p)^2

Simplify:

(1p)=9p    1=10p    p=110 (1-p) = 9p \implies 1 = 10p \implies p = \frac{1}{10}

Correct Answer: (B) 110 \frac{1}{10}

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Approach Solution -2

The probability mass function of a binomial distribution is: P(X=r)=(nr)pr(1p)nr. P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r}. Given P(X=2)=9P(X=3)P(X = 2) = 9P(X = 3), substitute into the formula: (52)p2(1p)3=9(53)p3(1p)2. \binom{5}{2} p^2 (1 - p)^3 = 9 \binom{5}{3} p^3 (1 - p)^2. Simplify the binomial coefficients: 542p2(1p)3=95436p3(1p)2. \frac{5 \cdot 4}{2} \cdot p^2 (1 - p)^3 = 9 \cdot \frac{5 \cdot 4 \cdot 3}{6} \cdot p^3 (1 - p)^2. 10p2(1p)3=90p3(1p)2. 10p^2(1 - p)^3 = 90p^3(1 - p)^2. Divide both sides by 10p2(1p)210p^2(1 - p)^2: 1p=9p. 1 - p = 9p. Simplify: 1=9p+p    1=10p    p=110. 1 = 9p + p \implies 1 = 10p \implies p = \frac{1}{10}.

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