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16 Sep 2014, 00:21
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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:
75%
(hard)
Question Stats:
55% (02:20) correct
45%
(02:35)
wrong
based on 732
sessions
History
Date
Time
Result
Not Attempted Yet
A shop had 10 bread loaves, 7 of which were baguettes. If every loaf has an equal chance of being sold, what is the probability that out of the 6 loaves sold, exactly 4 were baguettes?
A. \(\frac{2}{5}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{1}{2}\)
E. \(\frac{4}{7}\)
A. \(\frac{2}{5}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{1}{2}\)
E. \(\frac{4}{7}\)
16 Sep 2014, 00:21
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Official Solution:
A shop had 10 bread loaves, 7 of which were baguettes. If every loaf has an equal chance of being sold, what is the probability that out of the 6 loaves sold, exactly 4 were baguettes?
A. \(\frac{2}{5}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{1}{2}\)
E. \(\frac{4}{7}\)
The store had a total of 10 loaves of bread, consisting of 7 baguettes and 3 non-baguettes.
Out of the 6 loaves sold, we want to determine the probability that exactly 4 were baguettes and 2 were non-baguettes.
\(P(b=4) = \frac{C^4_7 * C^2_3}{C^6_{10}} = \frac{1}{2}\).
Where:
\(C^4_7\) - represents the number of ways to select 4 baguettes from the 7 available;
\(C^2_3\) - represents the number of ways to select 2 non-baguettes from the 3 available;
\(C^6_{10}\) - represents the total number of ways to choose 6 loaves from the 10 available.
Answer: D
A shop had 10 bread loaves, 7 of which were baguettes. If every loaf has an equal chance of being sold, what is the probability that out of the 6 loaves sold, exactly 4 were baguettes?
A. \(\frac{2}{5}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{1}{2}\)
E. \(\frac{4}{7}\)
The store had a total of 10 loaves of bread, consisting of 7 baguettes and 3 non-baguettes.
Out of the 6 loaves sold, we want to determine the probability that exactly 4 were baguettes and 2 were non-baguettes.
\(P(b=4) = \frac{C^4_7 * C^2_3}{C^6_{10}} = \frac{1}{2}\).
Where:
\(C^4_7\) - represents the number of ways to select 4 baguettes from the 7 available;
\(C^2_3\) - represents the number of ways to select 2 non-baguettes from the 3 available;
\(C^6_{10}\) - represents the total number of ways to choose 6 loaves from the 10 available.
Answer: D
General Discussion
01 Jul 2015, 18:03
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Bunuel
I obtained the same answer with a different numerator. I am curious if and why 6C4 is incorrect for numerator. If anyone has info, please kindly reply and thanks in advance!
