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The probability of rolling any number on a weighted 6-sided
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18 Sep 2019, 07:55
18 Sep 2019, 07:55
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The probability of rolling any number on a weighted 6-sided die, with faces numbered 1 through 6, is directly proportional to the number rolled. What is the expected value of a single roll of the die?
A)\(4{\frac{1}{6}}\)
B)\(4{\frac{1}{3}}\)
C)\(4{\frac{1}{2}}\)
D)\(4{\frac{2}{3}}\)
E)\(4{\frac{5}{6}}\)
A)\(4{\frac{1}{6}}\)
B)\(4{\frac{1}{3}}\)
C)\(4{\frac{1}{2}}\)
D)\(4{\frac{2}{3}}\)
E)\(4{\frac{5}{6}}\)
Re: The probability of rolling any number on a weighted 6-sided
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19 Sep 2019, 11:01
19 Sep 2019, 11:01
3
pranab01 wrote:
The probability of rolling any number on a weighted 6-sided die, with faces numbered 1 through 6, is directly proportional to the number rolled. What is the expected value of a single roll of the die?
A)\(4{\frac{1}{6}}\)
B)\(4{\frac{1}{3}}\)
C)\(4{\frac{1}{2}}\)
D)\(4{\frac{2}{3}}\)
E)\(4{\frac{5}{6}}\)
A)\(4{\frac{1}{6}}\)
B)\(4{\frac{1}{3}}\)
C)\(4{\frac{1}{2}}\)
D)\(4{\frac{2}{3}}\)
E)\(4{\frac{5}{6}}\)
Directly proportional so
1*1 + 2*2 + 3*3 + 4*4 + 5*5 + 6*6
where the first number represents the weight placed on each roll. We know the weights must sum to one to we normalize by dividing our expression by:
1 + 2 + 3 + 4 + 5 + 6 = 21
so we have:
(1 + 4 + 9 + 16 + 25 + 36)/21 = (55 + 36)/21 = 91/21 = 4 (7/21) = 4 1/3
Re: The probability of rolling any number on a weighted 6-sided
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21 Oct 2019, 21:59
21 Oct 2019, 21:59
1
could you explain it more clearly, I did not understand the first 2 steps you did
