Hello! I need help with the interpretation of this question as I believe this is without replacement, however the answers suggest that it is with replacement.

The ratio of girls to boys at a school is four to five. Two students are surveyed at random from the school. Find the probability that the students are

A) both boys

B) a girl and a boy

C) at least one girl.

The answers provided were

A) 25/81

B) 40/81

C) 56/81

I had the same problem with most of the questions from this exercise as by common sense I think it is without replacement but maybe I am reading the question wrong??

These questions have an extra subtlety behind them. CSSA papers love doing this but they try to make it a bit clearer.

\[ \text{The hidden assumption is that we're considering}\\ \text{a }\textbf{very large}\text{ sample.}\\ \text{Just for the sake of example, let's assume that the school}\\ \text{has 900 students.} \]

\[ \text{Then there would be 400 girls and 500 boys.}\\ \text{Suppose we want the probability of them both being boys.}\\ \text{Then the answer would be }\frac{500}{900}\times \frac{499}{899}. \]

In a similar way, let's say it has 1800 students (an arguably giant school). Then the probability would be \( \frac{1000}{1800} \times \frac{999}{1799}\).

\[ \text{Observe how technically speaking, the probability actually changes}\\ \text{a little when we change the number of students at the school.} \]

\[ \text{However if you plug into the calculator, you'll see that }\frac{499}{899} \approx \frac{500}{900} = 5/9\text{ anyway.}\\ \text{Similarly, }\frac{999}{1799}\approx \frac{1000}{1800} = \frac59. \]

Here's where everyone gets confused. People tend to think that because the ratio is \(4:5\), there should

**only** be \(9\) students. Thus they're lead to think that the required probability should be \( \frac{5}{9} \times \frac{4}{8} \).

When written in English, this makes no sense at all. But it's easy to miss this subtlety in the question because a lot of people forget that

**a ratio tells us nothing about how big the original sample is!**\[ \text{This is why we usually assume that the sample is very large.}\\ \text{When we do so, we see that after surveying one student}\\ \text{the ratio of remaining students isn't necessarily equal to }4:5\text{ anymore},\\ \text{but is still }\textbf{approximately}\text{ equal to }4:5. \]

\[\text{So we }\textbf{estimate}\text{ our required probability under this assumption.}\\ \text{This is why we arrive at }\frac59 \times \frac59\text{ instead.} \]

Note that these probabilities can never be computed to be exact. But at some point the error in the approximation is just negligible so we assume it doesn't matter. This question wasn't the best in my opinion because I feel uncomfortable assuming schools have 10000+ people, but it still illustrate the idea.