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Let's say I roll an unbalanced six-sided die where there's a 1/4 chance that I'll roll a six. I'm trying to find the probability distribution for how many times a 6 could appear if I roll the dice twice where X is a random variable that represents how many times a 6 could appear after rolling the die twice. Here's the table that I came up with:

   X | f(x)
   0 | 9/16
   1 | 6/16
   2 | 1/16

And E(X) = (0)(9/16) + (1)(6/16) + (2)(1/16) = 1/2

Now I also want to find the probability distribution for the average number of times that 6 could appear. The table I came up with for that is:

   x̄   | f(x̄)
   0   | 9/8
   0.5 | 15/16
   0.5 | 15/16
   1   | 5/8
   1   | 5/8
   1   | 3/4
   1.5 | 7/16
   1.5 | 7/16
   2   | 1/8

However my E(x̄) is 9/2 which can't be right because E(x̄) = E(X) and 9/2 doesn't equal 1/2. So where did I make my mistake.

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  • $\begingroup$ $9/16+3/16+1/16=13/16<1$ so this is not a correct probability distribution. The mistake is in this table. $3/16$ should be $6/16$ instead (by complementarity) $\endgroup$ – Jimmy R. Feb 22 '16 at 19:39
  • $\begingroup$ $9/8 > 1$, so it cannot be a probability. $\endgroup$ – Masacroso Feb 22 '16 at 20:00
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The probability of $X = 1$ is $$P(X = 1) = \binom{2}{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right) = \frac{6}{16}.$$ Then $$E[X] = 0 \cdot\frac{9}{16}+1\cdot\frac{6}{16}+2\cdot\frac{1}{16} = \frac{1}{2}.$$ Finally, $$E[\bar X] = \frac{1}{n}\cdot n E[X_1] = \frac{1}{2}.$$

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  • $\begingroup$ And are you also supposed to get 1/2 for the average because I got 9/2. Did I do my average calculations incorrectly? (I updated the question) $\endgroup$ – David Rolfe Feb 22 '16 at 19:54
  • $\begingroup$ You didn't clearly define what $\bar x$. If you are just taking the average of whether you got a six on a roll for two rolls, then the average cannot be greater than 1. $\endgroup$ – Em. Feb 22 '16 at 20:08
  • $\begingroup$ The average is the sampling distribution. That's what x̄ is. It's the sampling distribution of the mean. $\endgroup$ – David Rolfe Feb 22 '16 at 20:10
  • $\begingroup$ $$E[\bar X] = E\left[\frac{X_1+X_2}{2}\right] = \frac{1}{2}\left[E[X_1]+E[X_2]\right] = \frac{1}{2}\left[\frac{1}{2}+\frac{1}{2}\right] = \frac{1}{2}.$$ $\endgroup$ – Em. Feb 22 '16 at 20:13
  • $\begingroup$ I'm not sure what you did. To find the E(x̄) all I have to do is do what I did with E(X) and the two should match but when I do it with E(x̄) the two don't match so I'm wondering if there's a value in my table that's wrong. $\endgroup$ – David Rolfe Feb 22 '16 at 20:17

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