Q. 3.73

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.

(d) The $2$ oldest are girls.

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

Step-by-Step Solution

Verified

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, $P (E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4} E_{5}) = \frac{1}{32}$ .

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

d) The probability of the two oldest are girls, $P (E_{1} E_{2}) = \frac{1}{4}$ .

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

1Step 1: Given Information

Events:

$E_{i}$ - is the $i$ -th child is a girl

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

All events are independent.

2Step 2: Probability of each children were of comparable sex (Part a)

$P$ ("all boys or all girls") $= P (E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4}^{c} E_{5}^{c} \cup E_{1} E_{2} E_{3} E_{4} E_{5})$

$= P (E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4}^{c} E_{5}^{c}) + P (E_{1} E_{2} E_{3} E_{4} E_{5})$

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

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

$= \frac{1}{16}$

3Step 3: Probability of three oldest were boys, others were girls (Part b)

$P (E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4} E_{5}) = {(1 - \frac{1}{2})}^{3} {(\frac{1}{2})}^{2}$

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

$= \frac{1}{32}$

4Step 4: Probability of three were boys (Part c)

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

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

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

$= \frac{5}{32}$

5Step 5: Probability of the 2 aged were girls (Part d)