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29 Jan 2019, 03:02
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (02:08) correct
30%
(02:26)
wrong
based on 261
sessions
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Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 2 white balls. A ball is chosen at random from the first box and placed in the second box without observing its color. Then the ball is drawn from the second box. Find the probability that it is white.
A. 2/7
B. 3/8
C. 3/7
D. 4/7
E. 5/8
A. 2/7
B. 3/8
C. 3/7
D. 4/7
E. 5/8
29 Jan 2019, 03:58
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Bunuel
Probability of Last ball chosen white = Total White balls in Box 2 / Total balls in Box 2
Box 1: 3 red and 5 white balls
Box 2: 4 red and 2 white balls
Case 1: Ball chosen Box 1 is White and then Ball chosen from Box 2 is White
Favourable Probability = Probability of chosen ball white in box 1 * Probability of chosen ball white from box 2
i.e. Favourable Probability = (5/8)*(3/7)
Case 2: Ball chosen Box 1 is Red and then Ball chosen from Box 2 is White
Favourable Probability = Probability of chosen ball Red in box 1 * Probability of chosen ball white from box 2
i.e. Favourable Probability = (3/8)*(2/7)
Total Favourable Probability = Probability of case 1+ Probability of Case 2
i.e. Required Probability = (5/8)*(3/7) + (3/8)*(2/7) = (15+6)/56 = 21/56 = 3/8
Answer: Option B
Archit3110
Major Poster
29 Jan 2019, 04:09
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Bunuel
we can solve it considering two cases
case 1 : only white balls are chosen from box 1 and is transferred to box 2
P ( white ) ;5/8 * 3/7
case 2: red ball is chosen from box 1 and is transferred to box 2
P (white ) ; 3/8 * 2/7
5/8 * 3/7 + 3/8 * 2/7
15+6/56 = 21/56 ; 3/8
IMO B
