Q77 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
Bag A contains 3 white and 4 red balls, bag B contains 4 white and 5 red balls and bag $C$ contains 5 white and 6 red balls. If one ball is drawn at random from eath of these three bags, then the probability of getting one white and two red balls is
- A$\frac{268}{693}$
- B$\frac{310}{693}$
- C$\frac{38}{99}$
- Correct answerD$\frac{26}{63}$
Solution
Bag $A$ has 3 white and 4 red balls.
Bag $B$ has 4 white and 5 red balls. Bag $C$ has 5 white and 6 red balls. Case I : Probability that white ball is from Bag $A$ $=\frac{3}{7} \times \frac{5}{9} \times \frac{6}{11}=\frac{90}{693}$ Case II : Probability that white ball is from Bag $B$ $=\frac{4}{7} \times \frac{4}{9} \times \frac{6}{11}=\frac{96}{693}$
Case III : Probability that white ball is from Bag C $=\frac{4}{7} \times \frac{5}{9} \times \frac{5}{11}=\frac{100}{693}$
Required probability $=\frac{90}{693}+\frac{96}{693}+\frac{100}{693}=\frac{26}{63}$.
Bag $B$ has 4 white and 5 red balls. Bag $C$ has 5 white and 6 red balls. Case I : Probability that white ball is from Bag $A$ $=\frac{3}{7} \times \frac{5}{9} \times \frac{6}{11}=\frac{90}{693}$ Case II : Probability that white ball is from Bag $B$ $=\frac{4}{7} \times \frac{4}{9} \times \frac{6}{11}=\frac{96}{693}$
Case III : Probability that white ball is from Bag C $=\frac{4}{7} \times \frac{5}{9} \times \frac{5}{11}=\frac{100}{693}$
Required probability $=\frac{90}{693}+\frac{96}{693}+\frac{100}{693}=\frac{26}{63}$.
Hence, the correct answer is option: 'D'.