Question

The Patterson family has 3 kids, 1 boy and 2 girls. Suppose that for each birth, the probability of a boy birth is 1/2, and the probability of a girl birth is also 1/2. What is the fractional probability of having 1 boy and 2 girls, in any order, in a family's first 3 births?

Answer 1

Answer:

$Probability = \frac{1}{8}$

Step-by-step explanation:

Given

Represent Boys with B and Girls with G

$P(B) = \frac{1}{2}$

$P(G) = \frac{1}{2}$

Required

Find the probability or having 1 boy 2 girls

Since the order is not important, the probability is calculated as follows;

$Probability = P(B) * P(G) * P(G)$

Substitute $\frac{1}{2}$ for P(B) and P(G)

$Probability = \frac{1}{2} * \frac{1}{2} * \frac{1}{2}$

$Probability = \frac{1 * 1 * 1}{2 * 2 *2}$

$Probability = \frac{1}{8}$

Hence, the fractional probability is $\frac{1}{8}$