Parents keep having children until they have one girl, at which point they stop; and babies are girls with probability 0.49.
If we select a child uniformly at random (from the entire population of children), what's the probability he or she has exactly one sibling?
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I thought this would just be probability of BG 51/100*49/100 but it isn't.
P = P(G)P(BG| Choose G) + P(B)P(BG|choose B)
Probability to have exactly $1$ sibling $$P(Sibling \geq 1) = 0.51 $$ $$ P(Sibling \geq 2) = \sum_{i=0}^{\infty} (0.49)\cdot (0.51)^{i+2} = 0.49\cdot(0.51)^2 \cdot \frac{1}{1-0.51} = 0.51^2 $$ The probability of a family to have exactly two kids: $$0.51 - 0.51^2 = 0.51\cdot 0.49 = 0.2499\approx 25\% $$ Let $X_i$ be the event that all kids until $i-1$ are boys and kid $i$ is a girl Expected number of children per family: $$ E[children] = \sum_{i=1}^{\infty}i\cdot P(X_i) = \sum_{1}^{\infty}i\cdot (0.51)^{i-1}\cdot 0.49 = 0.49\times \sum_{1}^{\infty}i\cdot (0.51)^{i-1} $$
The infinite series sums up to $\frac{1}{0.49} \approx 2.04$. So we have $2.04$ expected number of children per family.
If you have N families, then the expected number of children is $2.04\cdot N$. So there are about $0.25\cdot N$ families who have exactly two kids, hence there are $2\cdot 0.25 \cdot N = 0.5 N$ children with exactly one sibling. So the probability is $\frac{0.5}{2.04}$
