Since each birth is independent from each other we can use the binomial distribution:
[tex]P\left(X=k\right)=\binom{n}{k}p^k\left(1-p\right)^{n-k}[/tex]In this case we have 6 births wich mean that n=6, the probability of success is 0.5. With this in mind let's calculate the probabilities we need.
a)
In this case we have:
[tex]\begin{gathered} P\left(X\ge3\right)=P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right) \\ =\binom{6}{3}\left(0.5\right)^3\left(1-0.5\right)^{6-3}+\binom{6}{4}\left(0.5\right)^4\left(1-0.5\right)^{6-4}+\binom{6}{5}\left(0.5\right)^5\left(1-0.5\right)^{6-5}+\binom{6}{6}\left(0.5\right)^6\left(1-0.5\right)^{6-6} \\ =0.6563 \end{gathered}[/tex]Therefore, the probability the had at least 3 girls is 0.6563
b)
In this case we have:
[tex]\begin{gathered} P\left(X\leq4\right)=1-P\lparen X>4) \\ =1-P\lparen X=5)-P\left(X=6\right) \\ =1-\binom{6}{5}\left(0.5\right)^5\left(1-0.5\right)^{6-5}-\binom{6}{6}\left(0.5\right)^6\left(1-0.5\right)^{6-6} \\ =0.8906 \end{gathered}[/tex]Therefore, the probability the had at most 4 girls is 0.8906
