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23 Jun 2019, 05:31
Kudos
Bookmarks
D
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:
60% (01:37) correct
40%
(01:41)
wrong
based on 10
sessions
History
Date
Time
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Not Attempted Yet
A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its color is observed and this ball along with two additional balls of the same color are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is
A) 1/5
B) 3/4
C) 3/10
D) 2/5
E) 1/4
A) 1/5
B) 3/4
C) 3/10
D) 2/5
E) 1/4
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23 Jun 2019, 06:11
Kudos
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As the question is worded, I think the answer is really "either 1/2 or 1/3, depending on the colour of the first ball". The first ball was "observed", so it's strange to treat it as an unknown. If this were a DS question, and you had, using both statements, all of the information in the question, would it be sufficient? If not, it can't be possible to solve it when it's a PS problem
I know what they mean though: if in advance you knew this procedure would be run (two new balls matching the first selection would be added to the bag), what's the probability the second selection is red? Then we have two cases:
- 2/5 of the time, the first selection is red. Then we'll be picking from 6 red and 6 black balls, so we'll have a 1/2 chance of picking a red the second time. So the probability of this precise sequence of events is (2/5)(1/2) = 1/5
- 3/5 of the time, the first selection is black. Then we'll be picking from 4 red and 8 black balls, so we'll have a 1/3 chance of picking a red the second time. So the probability of this precise sequence of events is (3/5)(1/3) = 1/5
Adding the results from each case, the answer is 2/5.
I know what they mean though: if in advance you knew this procedure would be run (two new balls matching the first selection would be added to the bag), what's the probability the second selection is red? Then we have two cases:
- 2/5 of the time, the first selection is red. Then we'll be picking from 6 red and 6 black balls, so we'll have a 1/2 chance of picking a red the second time. So the probability of this precise sequence of events is (2/5)(1/2) = 1/5
- 3/5 of the time, the first selection is black. Then we'll be picking from 4 red and 8 black balls, so we'll have a 1/3 chance of picking a red the second time. So the probability of this precise sequence of events is (3/5)(1/3) = 1/5
Adding the results from each case, the answer is 2/5.
08 Oct 2020, 07:24
Kudos
Bookmarks
\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)
Initially, the total number of balls = 4R + 6B = 10
There are 2 possible events here
(1) Pick a red and then put back 2 more reds (along with the one chosen) and choose a red ball from the 12 balls.
OR
(2) Pick a black and then put 2 more blacks (along with the one chosen) and choose a red ball from the 12 balls.
Case 1: P(Red) from 4 Red and 6 Black = \(\frac{4}{10} = \frac{2}{5}\)
Then Again P(Red) from 6 Red and 6 Black = \(\frac{6}{12} = \frac{1}{2}\)
The combined probability P(Red) and P(Red) = \(\frac{2}{5} * \frac{1}{2} = \frac{1}{5}\)
Case 2: P(Black) from 4 Red and 6 Black = \(\frac{6}{10} = \frac{3}{5}\)
Then P(Red) from 4 Red and 8 Black = \(\frac{4}{12} = \frac{1}{3}\)
The combined probability of P(Black) and P(Red) = \(\frac{3}{5} * \frac{1}{3} = \frac{1}{5}\)
Therefore the required probability = P(Case 1) + P(Case 2) = \(\frac{1}{5} + \frac{1}{5}= \frac{2}{5}\)
Option D
Arun Kumar
Initially, the total number of balls = 4R + 6B = 10
There are 2 possible events here
(1) Pick a red and then put back 2 more reds (along with the one chosen) and choose a red ball from the 12 balls.
OR
(2) Pick a black and then put 2 more blacks (along with the one chosen) and choose a red ball from the 12 balls.
Case 1: P(Red) from 4 Red and 6 Black = \(\frac{4}{10} = \frac{2}{5}\)
Then Again P(Red) from 6 Red and 6 Black = \(\frac{6}{12} = \frac{1}{2}\)
The combined probability P(Red) and P(Red) = \(\frac{2}{5} * \frac{1}{2} = \frac{1}{5}\)
Case 2: P(Black) from 4 Red and 6 Black = \(\frac{6}{10} = \frac{3}{5}\)
Then P(Red) from 4 Red and 8 Black = \(\frac{4}{12} = \frac{1}{3}\)
The combined probability of P(Black) and P(Red) = \(\frac{3}{5} * \frac{1}{3} = \frac{1}{5}\)
Therefore the required probability = P(Case 1) + P(Case 2) = \(\frac{1}{5} + \frac{1}{5}= \frac{2}{5}\)
Option D
Arun Kumar
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