Question

Two drawings$,$ each of $3$ balls$,$ are made from a bag containing $5$ white and $8$ black balls$,$ the balls not being replaced before the second trail$:$ find the chance that the first drawing will give $3$ white and the second $3$ black balls.

Answer 1

At the first trail$,$ $3$ balls may be drawn in ${}^{13}{C}_3$ ways$;$
and $3$ white balls may be drawn in ${}^5{C}_3$ ways$;$
Therefore the chance of $3$ white at first trail $=\frac{5.4}{1.2}÷\frac{\mathrm{13.12.11}}{\mathrm{1.2.3}}=\frac{5}{143}$.
When $3$ white balls have been drawn and removed$,$ the bag contains $2$ white and $8$ black balls$;$
therefore at the second trail $3$ balls may be drawn in ${}^{10}{C}_3$ ways $;$
and $3$ black balls may be drawn in ${}^8{C}_3$ ways$;$
Therefore the chance of $3$ black at the second trail $=\frac{\mathrm{8.7.6}}{\mathrm{1.2.3}}÷\frac{\mathrm{10.9.8}}{\mathrm{1.2.3}}=\frac{7}{15};$
Therefore the chance of the compound event $\frac{5}{143}\times \frac{7}{15}=\frac{7}{429}$.