Problem 68 - Algebra and Trigonometry Real Mathematics, Real People

Question

Assume that the probability of the birth of a child of a particular sex is \(50 \% .\) In a family with four children, what is the probability that (a) all the children are boys, (b) all the children are the same sex, and (c) there is at least one boy?

Step-by-Step Solution

Answer

The probabilities for (a) all children being boys is 0.0625, (b) all children being the same sex is 0.125, and (c) having at least one boy is 0.9375.

Step 1: Calculate the probability of all children being boys

We know that the probability of having a boy or a girl is 0.5 (50%). Since these are independent events, we will multiply the probabilities. \(0.5 * 0.5 * 0.5 * 0.5 = 0.0625\) which gives us the probability for all the children to be boys.

Step 2: Calculating the probability of all children being the same sex

This can be either all boys or all girls. We already calculated the probability for all boys in the previous step, which is 0.0625. The same probability applies for all girls as these are also independent events. We add these two probabilities up to get the total probability. \(0.0625 + 0.0625 = 0.125\) gives the probability that all children will be the same sex.

Step 3: Calculating the probability of at least one boy

It's easier to calculate the probability of the complementary event, which is the probability of all children being girls and then subtract this from 1. The probability of all children being girls is \(0.5 * 0.5 * 0.5 * 0.5 = 0.0625\), as calculated in Step 1. Substract this from 1 to get the probability of having at least one boy. So, \(1 - 0.0625 = 0.9375\) which represents the probability of at least one boy.