Question

A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made at random from the bag without replacements. Find the probability that the first draw yields 3 white balls and the second draw 3 red balls.

Answer 1

Let A : Event that 3 balls in the first draw are all white
B : Event that 3 balls in the second draw are all red.
3 balls can be drawn out 13 in ${}^{13}{C}_3$ and 3 white ball can be drawn out of 5 in ${}^5{C}_3$ ways.
$P\left(A\right)=\frac{{}^5{C}_3}{{}^{13}{C}_3}=\frac{5}{143}$
Since 3 balls are not replaced before the second draw, we are left with 8 red and 2 white balls.
Now, 3 balls can be drawn in ${}^{10}{C}_3$ ways and 3 red balls can be drawn in ${}^8{C}_3$ ways.
$P\left(B/A\right)=\frac{{}^8{C}_3}{{}^{10}{C}_3}=\frac{7}{15}$
$\therefore P(A\cap B)=P\left(A\right).P\left(B/A\right)=\frac{5}{143}.\frac{7}{15}=\frac{7}{429}$