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Question

A package of 6 fuses are tested where the probability an individual fuse is defective is 0.05.(That is ,5% of all fuses manufactured are defective.)a.What is the probability one fuse will be defective?b.What is the probability at least one fuse will be defective?c.What is the probability that more than one fuse will be defective,given that at least one is defective?

Answer 1

This is also a binomial random variable.

Part a is asking for P(X = 1).

P(X = 1) = (₆C₁) (0.05)¹ • (0.95)⁵ ≈ 0.2321.

Part b is asking for the probability that at least one is defective. P(X ≥ 1), means not 0. So if we find P(X = 0) and subtract it from 1, then we will have P(X ≥ 1).

So P(X = 0) = (₆C₀) (0.05)⁰ • (0.95)⁶ ≈ 0.7351, and therefore P(X ≥ 1) ≈ 1 - 0.7351 ≈ 0.2649.

Part c is asking for P(X > 1 | X ≥ 1).

For this, we need to invoke the conditional probability formula P(A | B) = ^P(A∩B)/_P(B). So the numerator must be both P(X > 1) and P(X ≥ 1), which is logically equivalent to P(X > 1). Then the denominator will just be P(X ≥ 1).

So the fraction will be ^{P(X > 1)}/_{P(X ≥ 1)}. We already know P(X ≥ 1) from part b, so it remains for us to find P(X > 1).

We're almost there. In the case of P(X > 1), we need to exclude both P(X = 0) and P(X = 1).

P(X > 1) = 1 - P(X = 0) - P(X = 1)

= 1 - (₆C₀) (0.05)⁰ • (0.95)⁶ - (₆C₁) (0.05)¹ • (0.95)⁵ ≈ 1 - 0.7351 - 0.2312 ≈ 0.0328.

P(X > 1 | X ≥ 1) = ^{P(X > 1)}/_{P(X ≥ 1)} = ^0.0328/_0.2649 ≈ 0.1237.