Question

Suppose that in a production of spark plugs the fraction of defective plugs has been constant at 2 % over a long time and that this process is controlled every half hour by drawing and inspecting two just produced. Find the probabilities of getting (a) no defectives, (b)1 defective, (c) 2 defectives. What is the sum of these probabilities?

Answer 1

Answer:

a) 96.04% probability of getting no defectives.

b) 3.92% probability of getting 1 defective.

c) 0.04% probability of getting 2 defectives.

You pick 2 plugs, so either you get no defective, 1 defective, or both defective. The sum of these probabilities must be 100%, which is what we get.

Step-by-step explanation:

For each plug, there are only two possible outcomes. Either it is defective, or it is not. The probability of a plug being defective is independent of other plugs. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

2 % over a long time and that this process is controlled every half hour by drawing and inspecting two just produced.

This means that $p = 0.02, n = 2$

(a) no defectives,

P(X = 0)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.02)^{0}.(0.98)^{2} = 0.9604$

96.04% probability of getting no defectives.

(b)1 defective,

P(X = 1)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{2,1}.(0.02)^{1}.(0.98)^{1} = 0.0392$

3.92% probability of getting 1 defective.

(c) 2 defectives.

P(X = 2)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.02)^{2}.(0.98)^{0} = 0.0004$

0.04% probability of getting 2 defectives.

What is the sum of these probabilities?

You pick 2 plugs, so either you get no defective, 1 defective, or both defective. The sum of these probabilities must be 100%, which is what we get.