Question

Um A contains 9 red balls and 11 white balls. Urn B contains 12 red balls and 3 white balls. A person is to roll a single fair die. If the result is a one or a two, then he is to randomly select a ball from urn A. Otherwise he is to randomly select a ball form urn B. The probability of obtaining a red ball, is

• A. $\frac{41}{60}$
• B. $\frac{19}{60}$
• C. $\frac{21}{35}$
• D. $\frac{35}{60}$

Answer 1

Answer: (A)

E: event of drawing a red ball;
$E_1=1\text{ or }2\text{ on die }\Rightarrow P\left(E_1\right)=\frac{1}{3}$
$E_2=3,4,5,6\text{ on die }\Rightarrow P\left(E_2\right)=\frac{2}{3}$
$E=\left(E\cap E_1\right)+\left(E\cap E_2\right)$
$P(E)=P\left(E_1\right)\cdot P\left(E/E_1\right)+P\left(E_2\right)P\left(E/E_2\right)$
Using the law of total probabilities,

$P(\text{ red ball })=\frac{2}{6}\cdot \frac{9}{20}+\frac{4}{6}\cdot \frac{12}{15}=\frac{41}{60}$