At least how many children should a family have so that, with probability 0.95, the family has at least 1 boy and at least 1 girl (boys and girls are equally likely).
What my teacher told me:
$$1-\{(1/2)^n +(1/2)^n\}> 0.95$$
Why can't we do it like this:
$$1-\sum_{k=n}^{\infty}\{(1/2)^k +(1/2)^k\} = 0.95$$ (where $n$ is required answer)
I am not able to identify the bug, or I think I have not understand the question completely.
Please help me.
Best Answer
$(1/2)^n+(1/2)^n$ is the probability of getting all boys are all girls if they have $n$ children. Your teacher is looking for the smallest $n$ for which this is less than $0.05$, so that the probability of having at least one child of each sex is greater than $0.95$. (The inequality should be non-strict: the smallest $n$ such that $(1/2)^n+(1/2)^n\le0.05$.)
Your $$\sum_{k\ge n}\left(\left(\frac12\right)^k+\left(\frac12\right)^k\right)$$
is hard to interpret at all. You’re adding the probability that all of the kids are of the same sex if they have $n$ kids, the probability that all are of the same sex if they have $n+1$ kids, and so on, but this sum of conditional probabilities has no obvious meaning. It’s not the probability that all of their kids are of the same sex if they have at least $n$ kids, if that’s what you were thinking: to get that, you’d have to have a probability distribution of number of kids, and you’d have to weight the $k$ term by the probability that they have $k$ kids. We have no such probability distribution, and indeed there can’t be one here, because they’re choosing the number of kids on the basis of what happanes if they have that many.