Conditional probability

Question

I have been working on a question for hours now and I thought I would ask stack exchange.

"Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose that 16% of all ticks in a certain location carry Lyme disease, 10% carry HGE, and 10% of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease?"

The answer is .236, and I was wondering if someone could tell me how that is. I'm convinced that the answer is .1 - "10% of the ticks that carry at least one of these diseases in fact carry both of them." So, what more information is needed than that sentence? The tick has HGE, and .1 of ticks that carry a disease carry another. So why isn't the answer simply .1? Thanks for your help in advance.

Answer 1

Your intuition is not unreasonable. Let $P(L|H)$ denote the probability that a tick with HGE also carries Lyme. But what "10% of the ticks that carry at least one of these diseases in fact carry both of them." actually tells you is that $P(L|H) \ge 10$%.

To see why, consider the venn diagrams below. The upper example is a rough sketch of your question. The pitfall is that the 10% denote the size of the intersection $P(L∧H)$ relative to the total area (blue and yellow) and not relative to all ticks. The conditional probability $P(L|H)$ is the ratio of the intersection of the circles (grey) and the yellow circle. Informally, a random tick with HGE will be somewhere in the yellow circle and might or might not carry Lyme. What we see is that the probability of a tick with HGE to also carry Lyme is higher then a Lyme tick having also HGE, just by comparing the size of the intersection to the size of the circles.

To make things clearer, consider the lower example. Assume P(L) = 0.3 and P(H) = 0.05 (30% of the ticks carry Lyme and 5% carry HGE). Let $P(L \land H) = 0.15 P(L \lor H)$, i.e. 15% of the ticks that carry at least one of the diseases carry both of them. Then, using the following formalizations you should find that $P(L|H)=91,3$%. $$P(L \lor H) = P(L) + P(H) - P(L \land H)$$ $$P(L \land H) = 0.15 \, P(L \lor H)$$

I hope this helps. I left the details for you, because working them out usually helps me to remember things better.