Question

A small school bus contains eight first-graders and ten second-graders. if two children are ill, what is the probability that at least one of them is a second-grader?

Answer 1

The probability is 83.95%.

Step-by-step explanation:

To find the probability that at least one of the two ill children is a second-grader, we need to calculate the probability that none of the ill children are second-graders and subtract that from 1.

Let's denote the probability of being a first-grader as P(1) and the probability of being a second-grader as P(2).

The probability that both ill children are first-graders is P(1) * P(1) = (8/18) * (7/17) = 0.1605.

So, the probability that at least one of the ill children is a second-grader is 1 - 0.1605 = 0.8395, or 83.95%.

Answer 2

at least means :
1) one of them OR
2) two of them


`\binom{10}{1} * \binom{8}{1} \: \: here \: we \: say \:`
we choose one of them from second graders and another one from first garders

that "or" which I mentioned means we have to use + (plus) to go to 2nd state that I've written

in 2nd we consider that
two of them are second graders in that case:

`\binom{10}{2} * \binom{8}{0} \:`
so at last we've got this

`\binom{10}{1} * \binom{8}{1} + \binom{10}{2} * \binom{8}{0} =`

you have to know the formula that I'm using