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A box contains exactly 24 balls, of which 12 are red and 12 are blue.
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Bunuel
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A box contains exactly 24 balls, of which 12 are red and 12 are blue. [#permalink] New post  03 Jul 2017, 01:07
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A box contains exactly 24 balls, of which 12 are red and 12 are blue. If two balls are to be picked from this box at random and without replacement, what is the probability that both balls will be red?

(A) 11/46
(B) 1/4
(C) 5/12
(D) 17/40
(E) 19/40
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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue. [#permalink] New post  03 Jul 2017, 01:17
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Bunuel wrote:
A box contains exactly 24 balls, of which 12 are red and 12 are blue. If two balls are to be picked from this box at random and without replacement, what is the probability that both balls will be red?

(A) 11/46
(B) 1/4
(C) 5/12
(D) 17/40
(E) 19/40



12C2 / 24C2 = 12*11/24*23 = 11/46

Answer: Option A
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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue. [#permalink] New post  03 Jul 2017, 09:46
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To choose 2 red balls among red balls : 12C2 = 6*11
To choose 2 balls out of 24 balls : 24C2 = 12*23

total probablity of choosing 2 red balls without replcement = 6*11/12*23 = 11/46
Ans :A
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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue. [#permalink] New post  03 Jul 2017, 10:20
Given data : 24 balls in the box - 12 red and 12 blue

To find : Probability that both balls are red

Denominator(2 balls chosen from 24 balls) \(24c2 = \frac{24*23}{2} = 12*23\)
Numerator(2 red balls from 12 available) \(12c2 = \frac{12*11}{2} = 6*11\)

Probability that 2 red balls are chosen(without replacement) = \(\frac{6*11}{12*23} = \frac{11}{46}\)(Option A)
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A box contains exactly 24 balls, of which 12 are red and 12 are blue. [#permalink] New post  04 Jul 2017, 08:58
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Bunuel wrote:
A box contains exactly 24 balls, of which 12 are red and 12 are blue. If two balls are to be picked from this box at random and without replacement, what is the probability that both balls will be red?

(A) 11/46
(B) 1/4
(C) 5/12
(D) 17/40
(E) 19/40

Straightforward:
\(\frac{favorable.outcome}{possible.outcomes}\) for 2 picks, then multiply.

Pick #1: 12 red balls. 24 balls total
Probability of red in 1st pick: \(\frac{12}{24}=\frac{1}{2}\)

Now there are 11 red and 23 total balls. (There's no replacement.)

Probability of red in 2nd pick: \(\frac{11}{23}\)

Multiply probability of both picks.
P(2 successive reds):
\((\frac{1}{2}*\frac{11}{23}) =\frac{11}{46}\)

Answer A
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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue. [#permalink] New post  09 Jul 2017, 09:35
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Simple probability:

probability first ball red: 12/24
probability second ball red: 11/23

total probability : 1/2*11/23 = 11/46
Option A

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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue. [#permalink] New post  09 Jul 2017, 18:20
\(\frac{12c2}{24c2} = \frac{12*11}{24*23} = \frac{11}{46}\). Ans - A.
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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue. [#permalink] New post  22 Jul 2020, 05:29
could someone give an explanation to why is this approach wrong and correct my thinking

approach 1. Selecting 2 red balls - 12c1 * 11c1
total - 24c1 * 23c1

instead of

approach 2. 12c2/24c2

although it comes up to the same thing however if i mix up 1 and 2 say 12c1*11c1/24c2 - it doesn't work.

I incorrectly took this approach and came up with 11/23 which is wrong but would like to know why we shouldn't mix these 2 approaches.
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