Exercise 6.1 #8
A royal family has children until it has a boy or until it has three children, whichever comes first. Assume that each child is a boy with probability 1/2. Find the expected number of boys in this royal family and the expected number of girls.
We will use the following general formula to find the expected number of boys and girls:
\(E(X)={ x }_{ 1 }\times P(X={ x }_{ 1 })+...+{ x }_{ k }\times P(X={ x }_{ k })\)
\(=\sum _{ i=1 }^{ k }{ { x }_{ I }P(X={ x }_{ i }) }\)
Let B be the numbr of boys and G be the number of girls. Additionally, we will assume that the probability of having a girl will be the compliment of the probability of having a boy, which will also be 1/2.
The problem states that the probability of having a boy is 1/2, thus the probability of the second child being a boy will be the probability of having a girl times the probability of having a boy (GB: 1/2 * 1/2), and the probability of the third child being a boy will be the probability of the first two being girls and the third being a boy (GGB: 1/2 * 1/2 *1/2). When we sum up the probabilities of the 1st 2nd and 3rd child being a boy we get the expected number of boys:
\(E(B)=(1)\frac { 1 }{ 2 } +(1)\frac { 1 }{ 4 } +(1)\frac { 1 }{ 8 } =\frac { 7 }{ 8 }\)
Since in each case we can only have one boy, the probabilities remain the same.
In order to get the expected number of girls, we will solve it a bit different than for boys, since the problem states that the royal family will have children until it is a boy or up to 3, it means they could have all 3 girls.
First we will get the probabilities of having 3 girls, 2 girls, 1 girl, multiply each probability by its outcome and finally add all of them together.
We find the following probability combinations for having girls and a boy (GGG, GGB, GB) and then multiply by the number of girls in each combination:
\(E(G)=(3)\frac { 1 }{ 8 } +(2)\frac { 1 }{ 8 } +(1)\frac { 1 }{ 4 } =\frac { 7 }{ 8 }\)