Question

A company makes automobile batteries. The probability of seeing a defected product \displaystyle{0.01}0.02. What is the probability of seeing 4 defected batteries in a sample of 200 batteries ?

Answer 1

Answer:

There is a 19.73% probability of seeing 4 defected batteries in a sample of 200 batteries.

Step-by-step explanation:

For each battery there are only two possible outcomes. Either they are defective, or they are not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

$n = 200, p = 0.02$

What is the probability of seeing 4 defected batteries in a sample of 200 batteries ?

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

$P(X = 4) = C_{200,4}.(0.02)^{4}.(0.98)^{196} = 0.1973$

There is a 19.73% probability of seeing 4 defected batteries in a sample of 200 batteries.