There are 15 plugs in a box, of which six are known to be faulty. Three plugs are taken at random from the box.
There are C(15,3) = 455 combinations of 3 plugs we can select from the 15.
Work out the probability that
(A) all three plugs are faulty
There are C(6,3) = 20 combinations of 3 plugs we can select from the 6 faulty
ones.
That's 20 out of 455 or 20/455 or 4/91 or only about 4.4% of the time.
Answer: 4/91sts of the time.
(B) none of the three plugs is faulty
Since there are 6 faulty plugs, and 20 plugs in all, there are 20-6=14 good
plugs. For none of the 3 to be faulty, all 3 must be good:
There are C(14,3) = 364 combinations of 3 plugs we can select from the 14 good
plugs.
That's 364 out of 455 or 364/455 or 4/5 or 80% of the time.
Answer: 4/5ths of the time.
(C) at least one of the three plugs is faulty.
When we see the words "at least one", we find the probability of the complement
event, and subtract that from 1.
What is the complement event? It's when we do not have any faulty ones at all.
We have already calculated that in (B).
So we subtract 4/5 from 1: 1-4/5 = 5/5-4/5 = 1/5, or 20% of the time.
Answer: 1/5th of the time.
Edwin