Question

An urn contains 6 white , 4 red and 9 black balls two balls are drawn at random find the probability that…..    i) Both balls are white.   ii) Both balls are red.

Answer 1

The urn contains 6 white , 4 red and 9 black balls.

The total number of balls is $6+4+9=19$.

We know that number of ways of choosing $m$ balls from $n$ different balls is $\frac{n!}{(n-m)!m!} ,0\leq m\leq n$.

Now,

The number of ways of choosing 2 ball from 19 balls is $\frac{19!}{2!17!} =171$.

The number of ways of choosing 2 white ball from 6 red balls is $\frac{6!}{2!4!} =15$

The number of ways of choosing 2 red ball from 4 red balls is $\frac{4!}{2!2!} =6$

Hence,

a) The probability that both balls are white is $\frac{15}{171}= \frac{5}{57}$.

b) The probability that both balls are red is $\frac{6}{171}= \frac{2}{57}$