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# A box contains exactly 24 balls, of which 12 are red and 12 are blue.

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A box contains exactly 24 balls, of which 12 are red and 12 are blue.  [#permalink]

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03 Jul 2017, 01:07
00:00

Difficulty:

5% (low)

Question Stats:

96% (00:59) correct 4% (01:11) wrong based on 83 sessions

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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]

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03 Jul 2017, 01:17
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

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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue.  [#permalink]

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03 Jul 2017, 09:46
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]

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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]

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04 Jul 2017, 08:58
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

1st pick: 12 red and 24 total balls, probability of red in 1st pick =

$$\frac{12}{24}$$ or $$\frac{1}{2}$$

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

Probability of picking red on the 2nd choice is $$\frac{11}{23}$$

P(2 successive reds) = $$(\frac{1}{2}$$ * $$\frac{11}{23})$$ = $$\frac{11}{46}$$

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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue.  [#permalink]

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09 Jul 2017, 09:35
1
Simple probability:

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

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

+1 kudos if u like the post.
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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue.  [#permalink]

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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]

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08 Feb 2019, 11:45
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Re: A box contains exactly 24 balls, of which 12 are red and 12 are blue.   [#permalink] 08 Feb 2019, 11:45
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