A Probability Problem about the $Trypanosoma$ parasite Considering the distribution of _Trypanosoma_ (parasite) lengths shown below, suppose we take a **sample of two** Tryoanosomes. What is the _probability_ that: a) **both** Trypanosomes will be shorter than 20um? b) **the first** Trypanosome will be **shorter than 20um** and the **second** will be **longer than 25um?** c) **exactly one** of the Trypanosomes will be **shorter than 20um** and **one** will be **longer than 25um**? !enter image description here I have no idea were to start. Even though it looks like an easy solution, this question is kind of confusing, especially the B and C part. Could someone please help me to get on the right track? Thank you for your help guys!

All three questions can be solved in a pretty similar fashion. The key to answering them is using the rule that, if $A$ and $B$ are independent events, then $P(A\cap B) = P(A)P(B)$ — in other words, multiply the probabilities of each event. So for example, the probability of selecting a parasite shorter than 20 micrometers is $0.01 + 0.34 = 0.35$. The probability of selecting two such parasites, then, is $(0.35)(0.35) = 0.1225$.

There is a twist in the final question. The second question dictates what order you select the parasites in. But the third question states simply that you pick two parasites. Therefore, you could pick the shorter parasite followed by the longer one or the longer one followed by the shorter one. If $A$ is the first event and $B$ the second, you're seeking $P(A\cup B)$. If $P(A\cap B) = 0$ — which is the case here; it's impossible to pick both the shorter parasite _and_ the longer one first — then $P(A\cap B) = P(A) + P(B)$.

Does that make sense?