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28 Jan 2021, 05:40
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (01:33) correct
26%
(02:02)
wrong
based on 96
sessions
History
Date
Time
Result
Not Attempted Yet
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
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 29/70
B. 5/12
C. 1/2
D. 31/70
E. 7/12
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
28 Jan 2021, 06:18
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Bunuel
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
03 Feb 2021, 23:52
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IMO A
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08 Feb 2021, 12:22
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Bunuel
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
