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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Probability

NDA 2025 · Mathematics · Probability

Let X be a random variable following binomial distribution with...

Let XX be a random variable following binomial distribution with parameters n=5n = 5 and p=kp = k. Further, P(X=1)=0.4096P(X = 1) = 0.4096 and P(X=2)=0.2048P(X = 2) = 0.2048. What is the value of kk?

Options

  1. A0.20.2
  2. B0.250.25
  3. C0.30.3
  4. D0.350.35

Correct answer

A. 0.20.2

Step-by-step solution

Given XX follows a binomial distribution with parameters n=5n = 5 and p=kp = k.

The probability mass function is given by P(X=x)=nCxpx(1−p)n−xP(X = x) = ^{n}C_{x} p^{x} (1-p)^{n-x}.

For x=1x = 1:
P(X=1)=5C1k1(1−k)4=5k(1−k)4=0.4096P(X = 1) = ^{5}C_{1} k^{1} (1-k)^{4} = 5k(1-k)^{4} = 0.4096

For x=2x = 2:
P(X=2)=5C2k2(1−k)3=10k2(1−k)3=0.2048P(X = 2) = ^{5}C_{2} k^{2} (1-k)^{3} = 10k^{2}(1-k)^{3} = 0.2048

Dividing P(X=1)P(X = 1) by P(X=2)P(X = 2), we get:
P(X=1)P(X=2)=5k(1−k)410k2(1−k)3\dfrac{P(X = 1)}{P(X = 2)} = \dfrac{5k(1-k)^{4}}{10k^{2}(1-k)^{3}}

0.40960.2048=1−k2k\dfrac{0.4096}{0.2048} = \dfrac{1-k}{2k}

2=1−k2k2 = \dfrac{1-k}{2k}

4k=1−k4k = 1 - k

5k=15k = 1

k=0.2k = 0.2
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