two numbers are selected at random from a set of first 90 natural numbers. find the probabilty that the product of randomly selected numbers is divisible by 3
Answers
Answer:
1/3
Step-by-step explanation:3*30=90
so 30 would be the probability
now 30/90 gives 1/3......
Answer:
5/9 = 0.555... (if repetition is allowed)
Step-by-step explanation:
0. I'll be proceeding by selecting the two numbers in an ordered way.
1. If the first number is divisible by 3, well, it doesn't matter what the second number is. The probability of this happening is: 1/3. (30 numbers in the first 90 which are divisible by 3)
2. Now, if, say, the first wasn't (whose chance is 2/3), the chances of the second number being so is also 1/3, since repetition is... enabled (let). The chance of this whole event happening is hence 2/3 * 1/3 = 2/9.
3. The total probability is 1/3 + 2/9 = 5/9.
The answer shall be different, and easy to solve but difficult to look at, if repetition wasn't allowed. In this case though, I'd directly assume it was.
Yes it is, but there are more cases: when the one number, and be careful about this, that we are talking about isn't divisible by 3, the other number, which we aren't talking about, might be. This is the extra probability which we aren't taking care of.