Probability with multipule selection?

Question

A factory has 6 machines,2 are modified, within a week problems are reported with 3 machines.

Assuming all machines are equally likley to give problems what is the probability that both of the modified machines are among the 3 with problems??


the answer is 0.20 i need to be shown how to get the answer

Answer 1

The probability is

(2 choose 2)(4 choose 1)/(6 choose 3)
= (1*4)/20 = 1/5 = 0.20

where (m choose n) = m!/[n! (m-n)!]

In words, the numerator is

Of the 2 modified machines choose 2, and of the 4 non-modified machines choose 1

This is the number of ways of choosing 3 machines, 2 of which are the 2 modified machines.

The denominator is the number of ways of choosing 3 machines out of a set of 6.