Chapter 3: Q. 3.73 (page 104)

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

Expert verified

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 $\frac{31}{32}$ .

Step by step solution

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.

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

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

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

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

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

$= \frac{1}{16}$

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

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

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

$= \frac{1}{32}$

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

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

$= \frac{5}{32}$

Probability of the 2 aged were girls (Part d)

P (E_{1} E_{2}) = {(\frac{1}{2})}^{2}

$= \frac{1}{4}$

Probability of at least one girl (Part e)

P

("at least one girl")

= 1 - P

(no girls)

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

$= \frac{31}{32}$

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Short Answer

Step by step solution

Given Information

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

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

Probability of three were boys (Part c)

Probability of the 2 aged were girls (Part d)

Probability of at least one girl (Part e)

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