Q:

Do I necessarily have to use conditional probability here?

There are $9$ girls and $10$ boys. I write names of all on slips of paper and put them in an urn. On the first day, I randomly pick out a slip and that student is the Class Representative for the day. Then that slip of paper is discarded. On the second day, I randomly pick out another slip and then, that student is Representative for the day. Since I discard the slips once I have picked them, no student can be a Representative more than once, till all the slips are exhausted.
If the first selection made is a boy, find the probability that the second selection is a girl.
Now, this problem is stated in words that clearly demand use of Conditional Probability. However, I thought of it this way:
Since I had selected a boy the previous day, that selection shouldn't matter the next day's selection. On the second day, I am left with $9$ girls and $9$ boys to pick from as the Representative. Out of them, I have to pick a girl. That gives me probability of $\frac{9}{18}=\frac12$
Am I thinking correctly?

A:

Yes that is correct. Lets verify this using conditional probability. Let $G_2$ = "second selection is a girl" and $B_1$ = "first selection is a boy". Then
$$\mathbb{P}(G_2|B_1) = \frac{\mathbb{P}(G_2 \cap B_1)}{\mathbb{P}(B_1)}$$
Now the only way the event $G_2\cap B_1$ can occur is if the first selection is a boy and the second selection is a girl. So if you draw a probability tree you will get the result $$\frac{\frac{10}{19}\times\frac{9}{18}}{\frac{10}{19}} = \frac{9}{18}=\frac{1}{2}$$

