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# A box contains 12 balls, seven of them are red and five of them are gr

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Math Expert
Joined: 02 Sep 2009
Posts: 58340
A box contains 12 balls, seven of them are red and five of them are gr  [#permalink]

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02 Oct 2018, 02:19
00:00

Difficulty:

25% (medium)

Question Stats:

78% (01:51) correct 22% (02:06) wrong based on 52 sessions

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A box contains 12 balls, seven of them are red and five of them are green. If three balls are to be selected at random from the box, what is the probability that two of the balls selected will be red and one will be green?

(A) 7/44

(B) 7/22

(C) 51/100

(D) 21/44

(E) 7/9

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Re: A box contains 12 balls, seven of them are red and five of them are gr  [#permalink]

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03 Oct 2018, 15:00
Bunuel wrote:
A box contains 12 balls, seven of them are red and five of them are green. If three balls are to be selected at random from the box, what is the probability that two of the balls selected will be red and one will be green?

(A) 7/44

(B) 7/22

(C) 51/100

(D) 21/44

(E) 7/9

Total possible outcomes = 12c3 = 220

Possible ways to select 2 red from 7
7c2 = 21

Possible ways to select 1 green from 5
5c1 = 5

(21*5)/220 = (21*5)(22*2*5) = 21/44

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Re: A box contains 12 balls, seven of them are red and five of them are gr  [#permalink]

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04 Oct 2018, 19:32
2
Bunuel wrote:
A box contains 12 balls, seven of them are red and five of them are green. If three balls are to be selected at random from the box, what is the probability that two of the balls selected will be red and one will be green?

(A) 7/44

(B) 7/22

(C) 51/100

(D) 21/44

(E) 7/9

The number of ways to select 2 red balls is 7C2 = (7 x 6)/2! = 21.

The number of ways to select 1 green ball is 5C1 = 5.

The total number of ways to select 3 balls from 12 is 12C3:

(12 x 11 x 10)/3! = (12 x 11 x 10)/(3 x 2) = 2 x 11 x 10 = 220

So the total probability is (21 x 5)/220 = 105/220 = 21/44.

Alternate Solution:

There are 3 different ways that one can draw 2 red (R) balls and 1 green (G) ball:
GRR or RGR or RRG

The probability of selecting GRR is 5/12 x 7/11 x 6/10 = 7/44

The probability of selecting RGR is 7/12 x 5/11 x 6/10 = 7/44

The probability of selecting RRG is 7/12 x 6/11 x 5/10 = 7/44

Since any one of these 3 outcomes satisfies the requirement, we see that the probability of selecting two red balls and one green ball is 7/44 + 7/44 + 7/44 = 21/44.

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Re: A box contains 12 balls, seven of them are red and five of them are gr   [#permalink] 04 Oct 2018, 19:32
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