A family has n children. We pick one of them at random and find out that she is a girl. Now, what is the probability that all n children are girls?
Solution from the textbook,
$$P(all\ girls) = \frac 1{2^n}$$
$$P(girl\ picked| all\ girls) = 1$$
$$P(girl\ picked) = \frac12$$
From Bayes' Theorem,
$$P(all\ girls| girl\ picked) = \frac{P(girl\ picked| all\ girls)*P(all\ girls)}{P(girl\ picked)}$$
$$P(all\ girls| girl\ picked) = \frac{1*\frac1{2^n}}{\frac12} = \frac{1}{2^{n-1}}$$
However, when I attempted it I did find the same probabilities for P(all girls) and P(girl picked|all girls) but did not get $P(girl\ picked) = \frac12$. Instead I summed over the probability that a girl was picked given there were i children(where $0\leq i\leq n$).
$$P(girl\ picked) = \sum_{i = 0}^{n} P(girl\ picked|i\ girls)P(i\ girls) $$
$$P(girl\ picked) = 0 + \frac 12 * \frac 1n + \frac {1}{2^2}\frac{2}{n} + \frac {1}{2^3}\frac{3}{n} + \frac {1}{2^4}\frac{4}{n} + …$$
$$P(girl\ picked) = \frac{1}{2n}\sum_{i = 0}^{n}\frac{i}{2^{i-1}} $$
Why is my interpretation of the probability that a girl was picked wrong. How did they arrive at $P(girl\ picked) = \frac{1}{2}$? Did they just assume that it's just as likely that a girl would be picked as it would a guy?
Best Answer
If you look at the sum approach you took, we are actually summing a binomial distribution: $P( i \text{ girls out of } n) = \binom{n}{i}\frac{1}{2^n}$ and the probability we pick a girl out of a family of $i$ girls and $n-i$ boys is $\frac{i}{n}$. You forgot about the binomial coefficients altogether!
So we're then left with $$P(\text{girl chosen from family of } n)=\sum_{i=0}^n \binom{n}{i}\frac{1}{2^n}\frac{i}{n} = \frac{1}{n}\sum_{i=0}^n i\binom{n}{i}\frac{1}{2^n} = \frac{1}{n}\frac{n}{2} = \frac{1}{2}$$ where the last sum is just the expectation of the binomial distribution, hence $\frac{n}{2}$ ($n$ trials with $\frac12$ success chance) and the $n$ drops out and we're left with $\frac{1}{2}$ as claimed.