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25 Jul 2019, 08:00
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In a group of 100, 25 drink only tea, 20 drink only coffee, and 15 drink only mocha. 5 drink none of the three and 10 drink all the three. What is the probability that a member chosen randomly drinks exactly two of the three drinks?
A. \(\frac{3}{20}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{10}\)
E. \(\frac{7}{20}\)
A. \(\frac{3}{20}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{10}\)
E. \(\frac{7}{20}\)
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25 Jul 2019, 08:23
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In a group of 100, 25 drink only tea, 20 drink only coffee, and 15 drink only mocha. 5 drink none of the three and 10 drink all the three. What is the probability that a member chosen randomly drinks exactly two of the three drinks?
A. 3/20
B. 1/5
C. 1/4
D. 3/10
E. 7/20
Please see solution in the image below
IMO C
A. 3/20
B. 1/5
C. 1/4
D. 3/10
E. 7/20
Please see solution in the image below
IMO C
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2019-07-25_205157.jpg [ 711.23 KiB | Viewed 6904 times ]
firas92
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25 Jul 2019, 09:37
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With reference to figure below, the required probability is \(\frac{(X+Y+Z)}{100}\)
Adding up the values from each region, we must get the total of 100
That is,
\((X+Y+Z)+10+25+20+15+5 = 100\)
\((X+Y+Z) = 25\)
Therefore, probability that a member chosen randomly drinks exactly two of the three drinks \(= \frac{25}{100} = \frac{1}{4}\)
Answer is (C)
Adding up the values from each region, we must get the total of 100
That is,
\((X+Y+Z)+10+25+20+15+5 = 100\)
\((X+Y+Z) = 25\)
Therefore, probability that a member chosen randomly drinks exactly two of the three drinks \(= \frac{25}{100} = \frac{1}{4}\)
Answer is (C)
Attachments
Untitled.png [ 33.22 KiB | Viewed 4738 times ]
