Question

In a factory there are 100100 units of a certain product, 55 of which are defective. We pick three units from the 100 units at random. What is the probability that exactly one of them is defective

Answer 1

Answer:

There is a 33.67% probability that exactly one of them is defective.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Here, we can have different formats. For example, D-ND-ND is the same as ND-D-ND, that is, the ordering is not important. So we use the combinations formula.

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.

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

Desired outcomes

One defective(one from a set of 55) and two non defective(two from a set of 45). So

$D = C_{55,1}*C_{45,2} = \frac{55!}{54!1}*\frac{45!}{43!2!} = 55*45*22 = 54450$

Total outcomes

Three from a set of 100. So

$T = C_{100,3} = \frac{100!}{97!3!} = 161700$

What is the probability that exactly one of them is defective

$P = \frac{D}{T} = \frac{54450}{161700} = 0.3367$

There is a 33.67% probability that exactly one of them is defective.