Probability of one vaccinated mother having a baby that didn't catch the flu, and another unvaccinated mother having a baby that caught the flu? A regional health board keeps records of the numbers of babies who get the flu when they are under six months old. It also records whether the mother had a flu vaccination during the six months before the baby was born. ![probability table]( If from among the group recorded in the above table, two babies and their mothers are chosen at random, then what is the probability that one mother was vaccinated and had a baby that didn't catch the flu, and the other was unvaccinated and had a baby that caught the flu? I tried to answer this question by adding 0.3511/0.3538 and 0.0863/0.6462, but could not get the correct answer of 0.0606. How do I solve this?

Consider the first mother being the one who was vaccinated and had a baby that didn't catch the flu (with a probability of $0.3511$) and the other mother being unvaccinated and had a baby that caught the flu (with a probability of $0.0863$). As these $2$ mothers, and babies, were picked at random, you can assume they are independent events, so the probability of this situation is the product of the probabilities, i.e.,

$$P_1 = 0.3511 \times 0.0863 = 0.03029993 \tag{1}\label{eq1}$$

However, as it's just as likely the situation occurred in the opposite order, i.e., the first mother was unvaccinated & the second one was vaccinated, then the probability of this occurring is also $P_1$, so the total probability would be

$$P_T = 2 \times P_1 = 0.060599986 \tag{2}\label{eq2}$$

Rounding $P_T$ to $4$ decimal places gives your stated correct answer of $0.0606$.