Question

Suppose that each child born to a couple is equally likely to be a boy or a girl, independently of the sex distribution of the other children in the family. For a couple having $5$children, compute the probabilities of the following events:

(a) All children are of the same sex.

(b) The $3$eldest are boys and the others girls.

(c) Exactly $3$are boys.

(d) The $2$oldest are girls.

(e) There is at least $1$girl.

Short answer

The probabilities of the following events:

a) The probability of All children are of the same sex is $\frac{1}{16}$

b) The probability of the three eldest are boys and the others girls, $P\left(E_1^c{E}_2^c{E}_3^c{E}_4{E}_5\right)=\frac{1}{32}$.

c) The probability of exactly three are boys is $\frac{5}{16}$.

d) The probability of the two oldest are girls, $P\left(E_1{E}_2\right)=\frac{1}{4}$.

e) The probability of there is at least one girl is $\frac{31}{32}$.

Step-by-step solution

Given Information

Events:

$E_i$- is the $i$-th child is a girl

$P\left(E_i\right)=\frac{1}{2}i=1,2,3,4,5...$

All events are independent.

Probability of each children were of comparable sex (Part a)

a)

$P$("all boys or all girls") $=P\left(E_1^c{E}_2^c{E}_3^c{E}_4^c{E}_5^c\cup E_1{E}_2{E}_3{E}_4{E}_5\right)$

$=P\left(E_1^c{E}_2^c{E}_3^c{E}_4^c{E}_5^c\right)+P\left(E_1{E}_2{E}_3{E}_4{E}_5\right)$

$={\left(1-\frac{1}{2}\right)}^5+{\left(\frac{1}{2}\right)}^5$

$={\left(\frac{1}{2}\right)}^4$

$=\frac{1}{16}$

Probability of three oldest were boys, others were girls (Part b)

b)

$P\left(E_1^c{E}_2^c{E}_3^c{E}_4{E}_5\right)={\left(1-\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^2$

$={\left(\frac{1}{2}\right)}^5$

$=\frac{1}{32}$

Probability of three were boys (Part c)

c)

Each defined distribution of girls/boys within a household has a ${\left(\frac{1}{2}\right)}^5$chance of occurring. This is frequently obvious from b), and every such distribution, like a), is mutually exclusive. There are $\left(\begin{array}{c}5\\ 3\end{array}\right)$events in which exactly three boys are present.

$P$("exactly three boys") $=\left(\begin{array}{l}5\\ 3\end{array}\right){\left(\frac{1}{2}\right)}^5$

$=10\times {\left(\frac{1}{2}\right)}^5$

$=\frac{5}{32}$

Probability of the 2 aged were girls (Part d)

d)

$P\left(E_1{E}_2\right)={\left(\frac{1}{2}\right)}^2$

$=\frac{1}{4}$

Probability of at least one girl (Part e)

e)

$P$("at least one girl") $=1-P$(no girls)

$=1-{\left(\frac{1}{2}\right)}^5$

$=\frac{31}{32}$