The probability of there being zero males is given by:
[tex]P(zero\ males)=4C0\times0.5^{0}\times0.5^{4}=\frac{1}{16}[/tex]
Similarly the probability of there being zero females is also 1/16.
The probability there will be at least 1 male and 1 female is therefore:
[tex]1-(\frac{1}{16}+\frac{1}{16})=\frac{7}{8}[/tex]
The answer is: 7/8.
