> My reasoning so far is from 1000 kids, 20 have chicken pox and 15 of those children have red spots. So then the answer would be 15/1000 = 3/200. I'm not sure that this address the third bullet point though.
Your reasoning is correct that: of $1000$ kids, $20$ kids are expected to have the pox and $15$ of these to show spots.
To address the third bullet point, you also need to note that we expect $980$ kids _not to_ have the pox and $9.8$ of these _to_ have spots (on average).
So of the $15+9.8$ kids who have red spots only $15$ have the pox. Thus the probability of having the chicken pox given red spots is:$$\frac{15}{24.8} = \frac{75}{124}$$
* * *
More formally: $$\begin{align} \mathsf P(C\mid R) & = \frac{\mathsf P(C\cap R)}{\mathsf P(C\cap R)+\mathsf P(\
eg C \cap R)} \\\\[1ex] & = \frac{2\%\times 75\%}{2\%\times 75\% + 98\%\times 1\%} \\\\[1ex] & = \frac{75}{124} \\\\[1ex] & \approx 60.5\% \end{align}$$