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Difficulty:
35%
(medium)
Question Stats:
72% (01:53) correct
28%
(02:13)
wrong
based on 560
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20 rolls of a fair six-sided die produced the following results:
What is the probability that one additional roll will increase the current average (arithmetic mean) score?
A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)
What is the probability that one additional roll will increase the current average (arithmetic mean) score?
A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)
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09 Aug 2012, 08:21
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Official Solution:
20 rolls of a fair six-sided die produced the following results:
What is the probability that one additional rolls will increase the current average (arithmetic mean) score?
A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)
The current mean score is \(\frac{1*4 + 2*3 + 3*5 + 4*2 + 5*2 + 6*4}{20} = \frac{67}{20} = 3.35\). Therefore, only a roll of 4, 5, or 6 will increase the mean score. Thus, the probability is \(\frac{3}{6}=\frac{1}{2}\).
Answer: C.
Hope it's clear.
20 rolls of a fair six-sided die produced the following results:
What is the probability that one additional rolls will increase the current average (arithmetic mean) score?
A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)
The current mean score is \(\frac{1*4 + 2*3 + 3*5 + 4*2 + 5*2 + 6*4}{20} = \frac{67}{20} = 3.35\). Therefore, only a roll of 4, 5, or 6 will increase the mean score. Thus, the probability is \(\frac{3}{6}=\frac{1}{2}\).
Answer: C.
Hope it's clear.
General Discussion
31 Mar 2024, 01:01
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Bunuel
How did you came to conclusion that only 4,5,6 will increase the mean ? what about 1 ,2,3 ?
