Questions: A couple plans to have three children. What is the probability that a) they have all boys? 1/8 b) they have at least one girl? 7/8

Solution Steps

For a single child, the probability of having a boy ($p$) is $0.5$ and the probability of having a girl ($q$) is also $0.5$. The results of the Bernoulli distribution analysis are as follows:

• Probability of success (having a boy): $$P(X = 1) = p = 0.5$$

• Probability of failure (having a girl): $$P(X = 0) = 1 - p = 0.5$$

• Mean ($\mu$): $$\mu = 1 \cdot p + 0 \cdot (1 - p) = p = 0.5$$

• Variance ($\sigma^2$): $$\sigma^2 = p \cdot (1 - p) = 0.25$$

• Standard Deviation ($\sigma$): $$\sigma = \sqrt{pq} = 0.5$$

To find the probability that the couple has all boys when they have three children, we use the binomial distribution formula: $$P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x}$$ where $n = 3$ (number of children), $x = 3$ (number of boys), $p = 0.5$, and $q = 0.5$.

Calculating this gives: $$P(X = 3) = \binom{3}{3} \cdot (0.5)^3 \cdot (0.5)^{0} = 1 \cdot 0.125 \cdot 1 = 0.125$$

To find the probability of having at least one girl, we can use the complement rule: $$P(\text{at least one girl}) = 1 - P(\text{all boys})$$ From the previous calculation, we have: $$P(\text{all boys}) = 0.125$$ Thus, $$P(\text{at least one girl}) = 1 - 0.125 = 0.875$$

Final Answer