Question

A bag contains 4 blue balls and 8 red balls If two balls are drawn from bag at random,what is the probability that one ball is blue and the other is red

Answer 1

There are 12 balls in the bag

2 balls out of 12 can be selected in $n = C_{12}^2$ ways.

One blue and one red ball can be selected in $m = C_4^1C_8^1$ ways.

Then the wanted probability is

$p = \frac{m}{n} = \frac{{C_4^1C_8^1}}{{C_{12}^2}} = 4 \cdot 8 \cdot \frac{{2! \cdot 10!}}{{12!}} = \frac{{4 \cdot 8 \cdot 2}}{{11 \cdot 12}} = \frac{{16}}{{33}}$

Answer: $\frac{{16}}{{33}}$