winnerhere wrote:Three numbers are selected from first 9 natural numbers.What is the probability that the product of those numbers are divisible by 10?
1) 3/4
2) 1/12
3) 11/42
4) 5/12
5) 7/18
If the product of three numbers between 1 and 9 inclusive will be a multiple of 10, we'll need to have prime factors of 2 and 5 in our product. That is, we'll certainly need to pick the '5', and we'll also need to pick at least one even number.
We have 9C3 = (9*8*7)/3! = 84 ways of choosing three numbers between 1 and 9, inclusive (the wording of the question is bad - I'm assuming the numbers need to be different). That will be the denominator of our probability.
Now, we certainly need to pick the '5' in order to get a product which is a multiple of 10. If we pick the 5, we then have 8C2 = (8*7)/2 = 28 ways to pick the other two numbers.
We aren't done, since this includes combinations like {1,5,7} which do not contain any even numbers. So there will be less than 28 ways to pick our three numbers (and notice that we now know our probability must be less than 28/84 = 1/3 -- there are only two answer choices which are possible, B and C, and B is far too small, so we could choose C and move on). We can count all of the selections that give us a '5' along with two odd numbers -- that is, we can count the selections we do *not* want -- and subtract this from 28; that will give us the number of ways to select the '5' and at least one even number. After we pick the '5', we can select two odd numbers from the four remaining odd numbers in 4C2 = 6 ways. So we can make 28-6 = 22 selections which include the '5' and at least one even number. Dividing by 84, the total number of possible selections, gives us the answer: 22/84 = 11/42.
