Calculate probability from a Negative Binomial Distribution with multiple success scenarios - 7th child born is either 3rd or 4th boy So the problem I am trying to solve states that, if a newborn is equally likely to be a boy or girl, what is the probability that 1) the third child is the first boy 2) the eighth child is the second girl and 3) the seventh child is the third OR fourth boy.
This seems to be a pretty straightforward Negative Binomial distribution problem, with X~NegBin(k,0.5) where k is the number of "successes," and I figured since they are equally likely, asking whether the Xth child is a boy or girl can be interchanged, so for numbers 1 and 2, they can just get plugged right into the equation for Neg. Bin. Distributions. I got 0.125 (1 * 0.5^3) for 1, and 0.02735 (7 * 0.5^8) for 2, but my question is with the changing success scenario on number 3.
I was thinking it might involve some sort of intersection operation on k=3 and k=4 for P(X=7), but I'm not sure that that makes sense, or if there is a better way to think about this problem.
Any tips on this are appreciated!

The events $A$: the seventh child is the third boy and $B$: the  seventh child is the fourth boy are disjoint. We want $\Pr(A\cup B)$, which is $\Pr(A)+\Pr(B)$.
We have $\Pr(A)=\binom{6}{2}(1/2)^7$ and $\Pr(B)=\binom{6}{3}(1/2)^7$. Add.
Note that we are not only assuming that at each birth boy and girl are equally likely, but also that the $8$ outcomes are independent.
