Family survey

A total of 160 families with 4 children each were surveyed with the following results.

Boys	4	3	2	1	0
Girls	0	1	2	3	4
Families	7	50	55	32	16

Let the probability of picking a boy or a girl be 1/2. Let P=boy, Q=girl.
What is the probability of getting:

a) All boys
b) three boys, one girl
c) two boys, two girls
d) one boy, three girls
e) All girls

Correct answer:

p₁ =  0.0438
p₂ =  0.3125
p₃ =  0.3438
p₄ =  0.2
p₅ =  0.1

Step-by-step explanation:

$$\begin{gathered} p_0=1/2=\frac{1}{2}=0.5 \\ n=160 \\ {p}_1=\frac{7}{n}=\frac{7}{160}=0.0438 \end{gathered}$$

$p_2=\frac{50}{n}=\frac{50}{160}=0.3125$

$p_3=\frac{55}{n}=\frac{55}{160}=0.3438$

$p_4=\frac{32}{n}=\frac{32}{160}=0.2$

$$\begin{gathered} p_5=\frac{16}{n}=\frac{16}{160}=\frac{1}{10}=0.1 \\ \text{ Verifying Solution: } \\ {P}_1=\left(\genfrac{}{}{0ex}{}{4}{4}\right)\cdot p_0^4\cdot (1-p_0{)}^{4-4}=(\genfrac{}{}{0ex}{}{4}{4})\cdot 0.5^4\cdot (1-0.5{)}^{4-4}=1\cdot 0.5^4\cdot 0.5^0=\frac{1}{16}=0.0625 \\ {P}_2=\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot p_0^3\cdot (1-p_0{)}^{4-3}=(\genfrac{}{}{0ex}{}{4}{3})\cdot 0.5^3\cdot (1-0.5{)}^{4-3}=4\cdot 0.5^3\cdot 0.5^1=\frac{1}{4}=0.25 \\ {P}_3=\left(\genfrac{}{}{0ex}{}{4}{2}\right)\cdot p_0^2\cdot (1-p_0{)}^{4-2}=(\genfrac{}{}{0ex}{}{4}{2})\cdot 0.5^2\cdot (1-0.5{)}^{4-2}=6\cdot 0.5^2\cdot 0.5^2=\frac{3}{8}=0.375 \\ {P}_4=\left(\genfrac{}{}{0ex}{}{4}{1}\right)\cdot p_0^1\cdot (1-p_0{)}^{4-1}=(\genfrac{}{}{0ex}{}{4}{1})\cdot 0.5^1\cdot (1-0.5{)}^{4-1}=4\cdot 0.5^1\cdot 0.5^3=\frac{1}{4}=0.25 \\ {P}_5=\left(\genfrac{}{}{0ex}{}{4}{0}\right)\cdot p_0^0\cdot (1-p_0{)}^{4-0}=(\genfrac{}{}{0ex}{}{4}{0})\cdot 0.5^0\cdot (1-0.5{)}^{4-0}=1\cdot 0.5^0\cdot 0.5^4=\frac{1}{16}=0.0625 \end{gathered}$$

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