Question

The Wilson family had 8 children. Assuming that the probability of a child being a girl is \( 0.5 \), find the probability that the Wilson family had at least 7 girls? at most 7 girls? Round your answ

Answer 1

The probability that the Wilson family had at least 7 girls is approximately 0.0352 or 3.52% rounded to two decimal places. The probability that they had at most 7 girls is approximately 99.61%.

Step-by-step explanation:

The probability of having a girl or a boy in a single birth event is \(0.5\) each. To find the probability of having a certain number of girls in 8 births, we can use the binomial probability formula. The probability of having exactly \(k\) girls in \(n\) births where the probability of success (having a girl) is \(p\) is given by the binomial probability formula:

`\[ P(X = k) = \binom{n}{k} * p^k * (1 - p)^(n - k) \]`

For the Wilson family, to find the probability of having at least 7 girls, we sum the probabilities of having 7, 8 girls:

`\[ P(\text{at least 7 girls}) = P(X = 7) + P(X = 8) \]`

Substituting the values into the formula:

`\[ P(X = 7) = \binom{8}{7} * 0.5^7 * 0.5^1 \]`

`\[ P(X = 8) = \binom{8}{8} * 0.5^8 * 0.5^0 \]`

Calculating these probabilities gives us the probability of having at least 7 girls. To find the probability of having at most 7 girls, we calculate the probability of having 0 to 7 girls and sum those probabilities:

`\[ P(\text{at most 7 girls}) = P(X = 0) + P(X = 1) + \dots + P(X = 7) \]`

After calculating the individual probabilities for each scenario, the final probability for at least 7 girls is approximately 0.0352 or (3.52%rounded to two decimal places. The probability for at most 7 girls is approximately 99.61%. This means that it's highly probable for the Wilson family to have at most 7 girls and significantly less likely to have at least 7 girls among their 8 children.