Probability: A Family of Five Children | College Algebra Review at MATHalino

Problem 1
A family with 5 children is selected at random, what is the probability that all are girls or all are boys?

Solution 1

Click here to expand or collapse this section

All are girls:
$P_G = \left( \dfrac{1}{2} \right)^5 = \dfrac{1}{32}$

All are boys:
$P_B = \left( \dfrac{1}{2} \right)^5 = \dfrac{1}{32}$

All are girls or all are boys:
$P = P_G + P_B = \dfrac{1}{32} + \dfrac{1}{32}$

$P = \dfrac{1}{16}$ ← answer

Problem 2
A family with 5 children is selected at random, what is the probability that there are 3 boys and 2 girls?

Solution 2

Click here to expand or collapse this section

$P = \left( \dfrac{1}{2} \right)^3 \left( \dfrac{1}{2} \right)^2 ({^5}C_3)$

$P = \dfrac{5}{16}$ ← answer

Note:

Among the 5 children, 3 are boys = ${^5}C_3$

among the 5 children, 2 are girls = ${^5}C_2$

in 5 children, 3 are boys and 2 are girls = $\dfrac{5!}{3! \, 2!}$

and that ${^5}C_3 = {^5}C_2 = \dfrac{5!}{3! \, 2!}$

Problem 3
A family with 5 children is selected at random, what is the probability that at least one child is a boy?

Solution 3

Click here to expand or collapse this section

$\displaystyle P = \sum_{x = 1}^5 \left[ \left( \dfrac{1}{2} \right)^x \left( \dfrac{1}{2} \right)^{5 - x} ({^5}C_x) \right]$

$P = \dfrac{31}{32}$ ← answer

Preferred Solution
$P = 1 - Q_{\text{all are girls}} = 1 - \left( \dfrac{1}{2} \right)^5$

$P = \dfrac{31}{32}$ ← answer

Navigation