Questions: If the probability of a child being a girl is 1/2, and a family plans to have 3 children, what are the odds against having all girls? The odds are to

Solution Steps

To find the probability of a family having all girls when they have 3 children, we use the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 3 \) (the number of trials, or children),
• \( x = 3 \) (the number of successes, or girls),
• \( p = \frac{1}{2} \) (the probability of success),
• \( q = 1 - p = \frac{1}{2} \) (the probability of failure).

Calculating this gives:

\[ P(X = 3) = \binom{3}{3} \cdot \left(\frac{1}{2}\right)^3 \cdot \left(\frac{1}{2}\right)^{3-3} = 1 \cdot \frac{1}{8} \cdot 1 = \frac{1}{8} = 0.125 \]

The odds against an event is calculated as the ratio of the probability of the event not occurring to the probability of the event occurring. Thus, the odds against having all girls is given by:

\[ \text{Odds against} = \frac{1 - P(X = 3)}{P(X = 3)} \]

Substituting the probability we found:

\[ \text{Odds against} = \frac{1 - 0.125}{0.125} = \frac{0.875}{0.125} = 7 \]

This means the odds against having all girls are \( 7 \) to \( 1 \).

Final Answer