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15 Jan 2025, 01:30
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C
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Difficulty:
25%
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Question Stats:
79% (01:26) correct
21%
(02:02)
wrong
based on 85
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A classroom set of calculators contains 10 graphing calculators and 26 four-function calculators. Serena needs to select 2 graphing calculators for students making up their trigonometry quiz. If she randomly chooses 2 calculators from the cabinet without replacement, what is the probability that both calculators will be graphing calculators?
A. \(\frac{1}{18}\)
B. \(\frac{5}{72}\)
C. \(\frac{1}{14}\)
D. \(\frac{1}{5}\)
E. \(\frac{337}{630}\)
A. \(\frac{1}{18}\)
B. \(\frac{5}{72}\)
C. \(\frac{1}{14}\)
D. \(\frac{1}{5}\)
E. \(\frac{337}{630}\)
15 Jan 2025, 02:30
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Selecting \(2\) graphing calculators randomly \(= ^{10}C_2\)
Required Probabilty \(= \dfrac{^{10}C_2}{^{(10+26)}C_2} = \dfrac{^{10}C_2}{^{(36)}C_2} = \dfrac{9\times 5}{18\times 35} = \dfrac{1}{14}\)
ANS C
Required Probabilty \(= \dfrac{^{10}C_2}{^{(10+26)}C_2} = \dfrac{^{10}C_2}{^{(36)}C_2} = \dfrac{9\times 5}{18\times 35} = \dfrac{1}{14}\)
ANS C
15 Jan 2025, 07:42
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You can do the 10 C 2 way.
Another way to think about this here is (10/36)*(10-1/(36-1)) = 10/140 = 1/14, C
Another way to think about this here is (10/36)*(10-1/(36-1)) = 10/140 = 1/14, C
