Three of the 15 books are defective. If two books are to be selected from the book, what is the probability that both two books are not defect?
P(both good) = P(1st selection is good) x P(2nd selection is good)sud21 wrote:Three of the 15 books are defective. If two books are to be selected from the book, what is the probability that both two books are not defect?
Aside
To begin, there are 12 good books out of 15, so P(1st selection is good) = 12/15
To find the second probability, we need to assume that the first selection was good. This means there are 14 books remaining, and 11 of them are good.
So, P(2nd selection is good) = 11/14
Now back to the probability...
P(both good) = P(1st selection is good) x P(2nd selection is good)
= (12/15)(11/14)
= 22/35
Cheers,
Brent
