Question

The quality assurance engineer of a receiving-sets manufacturer inspects receiving-sets in lots of 50. He selects 4 of the 50 receiving-sets at random and inspects them thoroughly. Assuming that 5 of the 50 receiving-sets in the current lot are defective, find the probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

Answer 1

Answer:

0.0430 = 4.30% probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

Step-by-step explanation:

Sets are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

In this question:

Lots of 50 means that $N = 50$

5 are defective, which means that $k = 5$

4 are selected, which means that $n = 4$

Find the probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

This is $P(X = 2)$. So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,50,4,5) = \frac{C_{5,2}*C_{45,2}}{C_{50,4}} = 0.0430$

0.0430 = 4.30% probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.