I was just confused if I am thinking this one right.
so the question is:
"In a set of numbers from 1 to 10. If two numbers are to be selected from these 10 numbers with replacement, then what is the probability that at least one of them is even?"
So I thought. there are four possible outcome,
and since 3 out of them contains even number So I thought the probability
was 3/4. Is this the right way of thinking this problem??
please help. thank you.
yes I have I know why you did 5 choose 2, because there are 5 even and 5 odd, and i know why you did 10 choose 2, because there are total of 10 numbers and you need to choose 2,
but What I don't get is why you did 1 minus, that whole rationalized combinatorics.
So I guess, I kind of can understand why you did each part of the equation. but I can't seem to understand why you have decided to put those numbers together.
actually the choose is incorrect, that's the hypergeometric situation which is correct when you are choosing WITHOUT replacement
This is binominal, since it is with replacement.
The 3/4 is correct, which can be achieved via 1-(1/2)(1/2) as well.