40.8k views
1 vote
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?

User Jujule
by
7.9k points

1 Answer

2 votes

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.

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) = (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.

User Ronelle
by
8.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories