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21 Sep 2020, 03:45
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C
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% (02:38) correct
30%
(02:43)
wrong
based on 209
sessions
History
Date
Time
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Not Attempted Yet
A box contains 5 white balls and 8 black balls. A ball is removed from the box and replaced by two balls of the other color (for example, if a white ball is removed, then it is replaced by two black balls and vise-versa). After that, a second ball is drawn. What is the probability that the second ball is black?
A. 1/2
B. 38/91
C. 53/91
D. 8/13
E. 5/7
Are You Up For the Challenge: 700 Level Questions
A. 1/2
B. 38/91
C. 53/91
D. 8/13
E. 5/7
Are You Up For the Challenge: 700 Level Questions
01 Feb 2021, 11:58
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Bunuel
P(2nd ball is black) = P(1st ball white AND 2nd ball black OR 1st ball black AND 2nd ball black)
= P(1st ball white AND 2nd ball black) + P(1st ball black AND 2nd ball black)
= P(1st ball white) x P(2nd ball black) + P(1st ball black) x P(2nd ball black)
= 5/13 x 10/14 + 8/13 x 7/14
= 50/182 + 56/182
= 106/182
= 53/91
Answer: C
Cheers,
Brent
General Discussion
21 Sep 2020, 04:47
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Case 1- If first ball removed is black
We have 7W and 7B
Probability of getting 1 Black is(8/13)*(7/14) = 28/91
Case 2- If first ball removed is white
We have 4W and 10B
Probability of getting 1 Black is (5/13)*(10/14)= 25/91
the probability that the second ball is black = 28/91 +25/91 = 53/91
We have 7W and 7B
Probability of getting 1 Black is(8/13)*(7/14) = 28/91
Case 2- If first ball removed is white
We have 4W and 10B
Probability of getting 1 Black is (5/13)*(10/14)= 25/91
the probability that the second ball is black = 28/91 +25/91 = 53/91
Bunuel
