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NaeemHasan
Joined: 06 Oct 2015
Last visit: 04 Feb 2019
Posts: 61
Given Kudos: 73
Location: Bangladesh
Concentration: Accounting, Leadership
Schools: Iowa State '21 Marshall '21 Stanford '21
Updated on: 24 Oct 2018, 08:59
Originally posted by NaeemHasan on 24 Oct 2018, 05:02.
Last edited by chetan2u on 24 Oct 2018, 08:59, edited 1 time in total.
Last edited by chetan2u on 24 Oct 2018, 08:59, edited 1 time in total.
Corrected the OA
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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:
15%
(low)
Question Stats:
83% (01:23) correct
17%
(01:48)
wrong
based on 227
sessions
History
Date
Time
Result
Not Attempted Yet
A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag, one after another, what is the probability that first ball is red and the second ball is yellow?
(A) 1/14
(B) 1/7
(C) 2/7
(D) 5/7
(E) 3/14
(A) 1/14
(B) 1/7
(C) 2/7
(D) 5/7
(E) 3/14
24 Oct 2018, 08:42
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NaeemHasan
We have 6R , 11Y & 5P
We have to pick 1st Red Ball so probability of that is \(\frac{6}{22}\)
Next we pick 1 Yellow ball , so the probability is \(\frac{11}{21}\)
Total probability is \(\frac{6}{22}\)*\(\frac{11}{21} = \frac{1}{7}\); Answer must be (B)
General Discussion
24 Oct 2018, 08:20
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NaeemHasan
So,I would approach this problem as below
R=6
Y=11
B=5
So 6C1/22C1 * 11C1/ 21C1
Note red= since we havewithdrawn one ball so net will be 22-1= 21 balls and now we have to pick yellow out of the total these 21 balls
(6*11 ) / (22*21) = 1/7
So OA should be 1/7
