Question

An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting
(i) 2 red balls
(ii) 2 blue balls
(iii) One red and one blue ball.

Answer 1

\[\text{ It is given that the urn contains 7 red and 4 black balls.}\]

\[\left( i \right) P\left( 2 \text{ red balls } \right) = \frac{7}{11} \times \frac{7}{11}\]

\[ = \frac{49}{121}\]

\[\left( ii \right) P\left( 2 \text{ blue balls } \right) = \frac{4}{11} \times \frac{4}{11}\]

\[ = \frac{16}{121}\]

\[\left( iii \right) P\left( \text{ one red ball and one blue ball} \right) = P\left( \text{ blue ball followed by red ball }\right) + P\left( \text{ red ball followed by blue ball}  \right)\]

\[ = \frac{4}{11} \times \frac{7}{11} + \frac{7}{11} \times \frac{4}{11}\]

\[ = \frac{28}{121} + \frac{28}{121}\]

\[ = \frac{56}{121}\]