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02 Oct 2018, 02:19
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
79% (01:58) correct
21%
(02:17)
wrong
based on 81
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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
(A) 7/44
(B) 7/22
(C) 51/100
(D) 21/44
(E) 7/9
03 Oct 2018, 15:00
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Bunuel
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
(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
Answer choice D.
Posted from my mobile device
04 Oct 2018, 19:32
Kudos
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Bunuel
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
(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.
Answer: D
