Exercise of conditional probability

Question

Consider a population made up of families in which there are more than five children, so that the percentages of families with 0, 1, 2, 3, 4 and 5 children are given by 5%, 15%, 30%, 25%, 20% and 5% respectively. Suppose further that the probability that a child is male is equal to 0.5. When selecting a random family from that population, what is the probability that all children are boys knowing that:

a) there is at least one male child?

b) the family has at least one child?

Answer 1

I would highly recommend doing b) first. It is far easier.

P(Family has at least one child) = P(1 Child) + P(2 Children) + P(3 Children) + P(4 Children) + P(5 Children) = 0.15+0.30+0.25+0.20+0.05=0.95

Alternatively, you can recognize that the family having at least one child is the same as the complement of the family having no children or 1-P(0 Children)=1-0.05=0.95

For a, you an assume gender is independent of the number of children. This allows you to multiply the probability of a child being male and the probability a family has x children. You have to be careful with this though. Here's how I would set it up:

P(1 Child AND Male) = 0.15 * 0.5 = 0.075

P(2 Children AND there is at least one male) = 0.30 (the probability of having two children) * 0.75 (the probability of having at least one male child if a family has two children)

0.75 can be found a multitude of different ways. I personally believe using complements will be the easiest. As we showed that the probability of having at 1 of something is 1-P(0 of something), we can re-use that same logic. In other words, the probability of a family having at least one male is equal to 1-Probability(All Female). The probability of having at least one male for a given number of children can be found in italics below:

So for 2 children, it would be 1-(0.5)² = 1-0.25 =0.75

For 3 children, it would be 1-(0.5)³ = 1-0.125 = 0.875

For 4 children, it would be 1-(0.5)⁴ = 1-0.0625 = 0.9375

For 5 children, it would be 1-(0.5)⁵ = 1-0.03125= 0.96875

So combining all of that would yield the following:

P(1 Child AND at least one male) = 0.15 * 0.5 = 0.075

P(2 Children AND at least one male) = 0.30 * 0.75 = 0.225

P(3 Children AND at least one male) = 0.25 * 0.875 = 0.21875

P(4 Children AND at least one male) = 0.20 * 0.9375 = 0.1875

P(5 Children AND at least one male) = 0.05 * 0.06875 = 0.0034375

Add up those 5 bolded numbers to get your answer!