One urn contains two black balls (labelled B1 and B2)

Solution:

Given that one urn contains two black balls and one white ball

and second urn contains one black ball and two white balls.

It is also given that one of the two urns is chosen, then a ball is randomly chosen from the urn, then second ball is chosen at random from the same urn without replacing the first ball

(a) Sample Space S = {B1B2, B1W, B2W, B2B1, WB1, WB2, W1W2, W1B, W2B, W2W1, BW1, BW2}

Total number of sample space = 12

(b) If two black balls are chosen

Total outcomes = 12

Favourable outcomes are B1B2, B2B1

∴ Total favourable outcomes = 2

We know that,

Probability $=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$

$\therefore$ Required Probability $=\frac{2}{12}=\frac{1}{6}$

(c) If two balls of opposite colours are chosen

 

Favourable outcomes are $\mathrm{B}_{1} \mathrm{~W}, \mathrm{~B}_{2} \mathrm{~W}, \mathrm{WB}_{1}, \mathrm{WB}_{2}, \mathrm{~W}_{1} \mathrm{~B}, \mathrm{~W}_{2} \mathrm{~B}, \mathrm{BW}_{1}, \mathrm{BW}_{2}$

$\therefore$ Total favourable outcomes $=8$ and total outcomes $=12$

We know that,

Probability $=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$

$\therefore$ Required Probability $=\frac{8}{12}=\frac{2}{3}$