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A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two b

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A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two b  [#permalink]

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New post Updated on: 24 Oct 2018, 08:59
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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

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.
Corrected the OA
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Re: A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two b  [#permalink]

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New post 24 Oct 2018, 08:20
NaeemHasan wrote:
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



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


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Re: A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two b  [#permalink]

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New post 24 Oct 2018, 08:42
NaeemHasan wrote:
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


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)
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Re: A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two b  [#permalink]

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New post 24 Oct 2018, 10:08
The OA is given 2/7. Total number of ways to select 2 balls is 22C2. The number of ways to select red ball is 6C1 and the number of ways to select yellow ball is 11C1.
Hence, probability = 6C1*11C1/22C2= 2/7
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Re: A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two b  [#permalink]

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New post 25 Oct 2018, 17:52
NaeemHasan wrote:
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



We are not replacing the first ball after it is drawn. Thus, the probability is:

6/22 x 11/21 = 3/11 x 11/21 = 3/21 = 1/7.

Answer: B
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Re: A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two b   [#permalink] 25 Oct 2018, 17:52
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