There are two bags A and B. Bag A contains 2 white and 3 black balls.

Question

There are two bags A and B. Bag A contains 2 white and 3 black balls. Bag B contains 3 white and 4 black balls. A bag is chosen at random. What is the probability that the ball drawn is white?

A. 29/70
B. 5/12
C. 1/2
D. 31/70
E. 7/12

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GMATinsight · Answer

Bag A contains 2 white and 3 black balls
Bag B contains 3 white and 4 black balls

Case 1: Bag chosen is Bag A

Probability of Bag A to be chosen  = \frac{1}{2} 
Probability of White ball being chosen from Bag A  = \frac{2}{5}

i.e. Probabilty that Case 1 happens  = (\frac{1}{2})*(\frac{2}{5}) = \frac{1}{5}

Case 2: Bag chosen is Bag B

Probability of Bag B to be chosen  = \frac{1}{2} 
Probability of White ball being chosen from Bag B  = \frac{3}{7}

i.e. Probabilty that Case 2 happens  = (\frac{1}{2})*(\frac{3}{7}) = \frac{3}{14}

Required Probability = Case 1 or Case 2  = \frac{1}{5} + \frac{3}{14} = \frac{29}{70}

Answer: Option A

Harsh9676 · Answer

IMO A

ScottTargetTestPrep · Answer

Solution:
 
P(white ball) = P(white ball from bag A) * P(bag A) + P(white ball from bag B) * P(bag B)

= 2/5 * 1/2 + 3/7 * 1/2 
 
 = 1/5 + 3/14 = 14/70 + 15/70 = 29/70

Answer: A

There are two bags A and B. Bag A contains 2 white and 3 black balls.

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There are two bags A and B. Bag A contains 2 white and 3 black balls.

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