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,

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