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

Question:

One urn contains two black balls (labelled B1 and B2) and one white ball. A second urn contains one black ball and two white balls (labelled W1 and W2).

Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is

chosen at random from the same urn without replacing the first ball.

(a) Write the sample space showing all possible outcomes

(b) What is the probability that two black balls are chosen?

(c) What is the probability that two balls of opposite colour are chosen?

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 $=\frac{8}{12}=\frac{2}{3}$

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