Theoretical DistributionsPYQ Jan 26Question 4243 of 230
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A quality control inspector finds that 20% of light bulbs are defective. If a batch of 5 light bulbs is tested, what is the probability that exactly 1 bulb is defective?

Options

A0.4096
B0.8026
C0.2746
D0.1296
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Correct Answer

✅ Option a — 0.4096

All Options:

  • A0.4096
  • B0.8026
  • C0.2746
  • D0.1296

Detailed Solution & Explanation

This problem can be modeled using the Binomial Distribution formula:
P(X=x)=(nx)pxqn−xP(X = x) = \binom{n}{x} p^x q^{n-x}
where:
- n=5\displaystyle n = 5 (the number of trials/bulbs tested).
- p=20%=0.2\displaystyle p = 20\% = 0.2 (probability of success, i.e., a bulb being defective).
- q=1−p=1−0.2=0.8\displaystyle q = 1 - p = 1 - 0.2 = 0.8 (probability of failure, i.e., a bulb being non-defective).
- x=1\displaystyle x = 1 (we want to find the probability of exactly 1\displaystyle 1 defective bulb).

Substitute the values into the formula:
P(X=1)=(51)(0.2)1(0.8)5−1P(X = 1) = \binom{5}{1} (0.2)^1 (0.8)^{5-1}
P(X=1)=5×0.2×(0.8)4P(X = 1) = 5 \times 0.2 \times (0.8)^4
P(X=1)=1×(0.8)4P(X = 1) = 1 \times (0.8)^4

Now calculate (0.8)4\displaystyle (0.8)^4:
(0.8)2=0.64(0.8)^2 = 0.64
(0.8)4=0.64×0.64=0.4096(0.8)^4 = 0.64 \times 0.64 = 0.4096
Thus, P(X=1)=0.4096\displaystyle P(X = 1) = 0.4096.

Hence, **Option A** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

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For a Poisson distribution,

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