Question

A box contains 2 red, 3 black and 5 white balls. If 3 balls are drawn at a random without replacement, find the probabilities that:
2 are red and 1 black

Answer 1

2R+3B+5W
Total number of balls = 10
Number of combinations when 3 balls are drawn at random =${}^{10}{C}_3=\frac{10\times 9\times 8}{3\times 2\times 1}=120$
Number of ways 2 Red balls can be picked from 2 Red balls =${}^2{C}_2=1$
Number of ways 1 black can be picked up from 3 black =${}^3{C}_1=3$
$\therefore Numberofways2R+1B\to 1\times 3=3$
$\therefore Requiredprobably=\frac{3}{120}=\frac{1}{40}$