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16 Sep 2014, 00:22
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A box contains 60 pieces of candy in red, green, blue, and white colors. If a candy is drawn at random from the box and the probability that it is white is \(\frac{1}{5}\), and the probability that it is green is \(\frac{1}{4}\), what is the probability that the candy will be either red or blue?
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{9}{20}\)
D. \(\frac{11}{20}\)
E. \(\frac{19}{20}\)
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{9}{20}\)
D. \(\frac{11}{20}\)
E. \(\frac{19}{20}\)
16 Sep 2014, 00:22
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Official Solution:
A box contains 60 pieces of candy in red, green, blue, and white colors. If a candy is drawn at random from the box and the probability that it is white is \(\frac{1}{5}\), and the probability that it is green is \(\frac{1}{4}\), what is the probability that the candy will be either red or blue?
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{9}{20}\)
D. \(\frac{11}{20}\)
E. \(\frac{19}{20}\)
The probability of drawing a White or Green candy is \(\frac{1}{5} + \frac{1}{4} = \frac{9}{20}\).
Consequently, the probability of drawing either a Red or Blue candy is \(1 - \frac{9}{20} = \frac{11}{20}\).
Answer: D
A box contains 60 pieces of candy in red, green, blue, and white colors. If a candy is drawn at random from the box and the probability that it is white is \(\frac{1}{5}\), and the probability that it is green is \(\frac{1}{4}\), what is the probability that the candy will be either red or blue?
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{9}{20}\)
D. \(\frac{11}{20}\)
E. \(\frac{19}{20}\)
The probability of drawing a White or Green candy is \(\frac{1}{5} + \frac{1}{4} = \frac{9}{20}\).
Consequently, the probability of drawing either a Red or Blue candy is \(1 - \frac{9}{20} = \frac{11}{20}\).
Answer: D
General Discussion
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
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13 Jul 2015, 05:19
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Bunuel
Given, R+G+B+W = 60.
Now, P(w) = 1/5 ----> Number of white candies = 1/5 * 60 = 12
Similarly, P(G) =1/4 ----> Number of green candies = 1/4 * 60 = 15
Total, white + green = 12+15 = 27.
Remaining = 60 -27 = 33 = Red + Blue
Thus Probability of a candy being red or blue = P(R +B)/ Total = 33/60 = 11/20. Thus D is the correct answer.
