Respuesta :
Solution:
Let white balls be w, and black balls be b;
In the first bag;
[tex]n(w)=5,n(b)=3,n(T)=8[/tex]But, in the second bag;
[tex]n(w)=4,n(b)=6,n(T)=10[/tex](i) The probability that both balls drawn from each bag are white is;
[tex]\begin{gathered} P(w\text{ and }w)=\frac{5}{8}\times\frac{4}{10} \\ P(w\text{ and }w)=\frac{20}{80} \\ P(w\text{ and }w)=\frac{1}{4} \end{gathered}[/tex](ii) The probability that both balls drawn from each bag are black is;
[tex]\begin{gathered} P(b\text{ and }b)=\frac{3}{8}\times\frac{6}{10} \\ P(b\text{ and }b)=\frac{18}{80} \\ P(b\text{ and }b)=\frac{9}{40} \end{gathered}[/tex](iii) The probability that one is white and another is black is;
[tex]\begin{gathered} P(w\text{ and }b)=(\frac{5}{8}\times\frac{6}{10})+(\frac{3}{8}\times\frac{4}{10}) \\ P(w\text{ and }b)=\frac{30}{80}+\frac{12}{80} \\ P(w\text{ and }b)=\frac{42}{80} \\ P(w\text{ and }b)=\frac{21}{40} \end{gathered}[/tex]
