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

Transcript text: A couple plans to have three children. What is the probability that a) they have all boys? $\frac{1}{8}$ b) they have at least one girl? $\frac{7}{8}$

Solution

Solution Steps

Step 1: Bernoulli Distribution Analysis

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 ($\sigma$): \[ \sigma = \sqrt{pq} = 0.5 \]

Step 2: Probability of Having All Boys

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: \[ P(X = 3) = \binom{3}{3} \cdot (0.5)^3 \cdot (0.5)^{0} = 1 \cdot 0.125 \cdot 1 = 0.125 \]

Step 3: Probability of Having at Least One Girl

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, \[ P(\text{at least one girl}) = 1 - 0.125 = 0.875 \]