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16 Sep 2014, 01:22
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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. \(\frac{5}{6}\)
16 Sep 2014, 01:22
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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 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. \(\frac{5}{6}\)
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
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. \(\frac{5}{6}\)
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
15 Oct 2023, 15:45
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
