If 10 pairs of fair dice are rolled, approximate the probability that the sum of the values obtained (which ranges from 20 to 120) is between 30 and 40 inclusive.

I dont know where to start with this one. I have been looking all over the web for example, but nothing i find is applicable for finding the sum of numbers.

any advice would be great

  • $\begingroup$ Start by listing the combinations of numbers that result in a sum between 30 and 40. $\endgroup$ – Eckhard Apr 5 '13 at 16:06
  • $\begingroup$ i dont think so man, the combination would be almost endless. $\endgroup$ – KevinCameron1337 Apr 5 '13 at 16:58
  • $\begingroup$ That depends on one's understanding of almost endless. I admit that I overlooked the word 'approximate' in your question, though. $\endgroup$ – Eckhard Apr 5 '13 at 17:05

We are tossing $20$ dice. Let $X_i$ be the number showing on the $i$-th die. We are interested in the random variable $X_1+\cdots+X_{20}$.

The $X_i$ are independent, mean $\frac{7}{2}$, variance $\frac{35}{12}$ (please verify).

So $Y$ has mean $\mu=20\cdot\dfrac{7}{2}$, variance $\sigma^2=20\cdot \dfrac{35}{12}$.

We cross our fingers and use the normal approximation. This is because $Y$ is the sum of a not terribly small number of independent identically distributed respectable random variables $X_i$.

So we want the probability that if $W$ is normal with mean $\mu$ and variance $\sigma^2$, then $30\le W\le 40$.

The rest depends to some degree on whether you are expected to use the continuity correction.

Without the continuity correction, we want the probability that a standard normal $Z$ satisfies $\frac{30-\mu}{\sigma}\le Z\le \frac{40-\mu}{\sigma}$.

With continuity correction, replace $40$ by $40.5$, and $35$ by $34.5$.

If you have any difficulty finishing, please leave a comment.

  • $\begingroup$ i never heard of "continuity correction" so i guess not. Im working on it all now, thanks. $\endgroup$ – KevinCameron1337 Apr 5 '13 at 16:20
  • $\begingroup$ When we are using the normal to approximate, for example, a binomial $X$, the probability that $X\le k$ is often better approximated by the probability that $W\le k+\frac{1}{2}$, where $W$ is the normal with same mean and variance. (The no continuity correction uses $\Pr(W\le k)$.) Makes no significant difference if $n$ is large, like $400$. But for smallish $n$, the correction often leads to improved estimates. If your course has not discussed it, you are not expected to use it! I was just being cautious. $\endgroup$ – André Nicolas Apr 5 '13 at 16:26
  • $\begingroup$ So here is what i got P(z <= 40 - 70 / 175/3) = .303 and P(z <= 30 - 70 / 175/3) = .246. So, .303-.246 = 0.057 $\endgroup$ – KevinCameron1337 Apr 5 '13 at 16:31
  • $\begingroup$ Should the Variance be $sqrt(20)*35/12 $\endgroup$ – KevinCameron1337 Apr 5 '13 at 16:43
  • $\begingroup$ The standard deviation of $Y$ is about $7.6376$. The prob. that $Y\le 40$ is roughly the probability that $Z\le \frac{40-70}{7.6376}$. This is virtually $0$. I suspect there is a typo in the problem. $\endgroup$ – André Nicolas Apr 5 '13 at 16:43

Let's $x_1,\ldots x_{20}\in\{1,2,3,4,5,6\}$. For a exact solution $$ P\left(30\leq x_1+\ldots+x_{20}\leq 40 \right)=\frac{1}{20^6}\sum_{30\leq N\leq 40}\quad\sum_{x_1+\ldots+x_{20}\leq N} \frac{N!}{x_1!\cdot \ldots \cdot x_{10}!} $$


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