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When calculating the $P$ for at least $x$ success one uses $\text{max} (x-1)$ instead, and then take $1- (\text{max} (x-1))$. This works. And I understand it. Because we use the complement to calculate it, because the calculator supports it. But what I do not understand is the following. When calculating a combination of these, $P(\text{max}\,\, x\,\,\, \text{and}\,\,\, \text{min}\,\, y)$ we can just forget about the $1 - (\text{max}\,\, (x-1))$ part, and just use $\text{max}\,(x-1)$ directly. For example: $$P(\text{at least 150 sixes and at most 180 sixes)} = P(\text{max}\,\, 180 \,\,\text{sixes}) - P(\text{max}\,\,149\,\,\text{sixes}).$$ And then we don't have to do the $1-x$ part. Why is this?

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@Henry, I imagine what's meant is someone is throwing dice and is interested in the probability of a 6 turning up at least 150 times but at most 180 times. –  Gerry Myerson Jun 26 '11 at 0:54

3 Answers 3

up vote 1 down vote accepted

If you threw 1000 dice, you might want to know $$\Pr(\text{at least 150 sixes and at most 1000 sixes)} = \Pr(\text{at most 1000 sixes}) - \Pr(\text{at most 149 sixes}).$$

But you cannot get more than 1000 sixes from 1000 dice, so $\Pr(\text{at most 1000 sixes}) =1$, and you can rewrite this more briefly as $$\Pr(\text{at least 150 sixes)} = 1 - \Pr(\text{at most 149 sixes}).$$

In other words, the method in you first case is a particular of the method in your second case.

Incidentally, by the time you get to 150 sixes you could be using the central limit theorem, in which case you are using "max" because many tables and calculators give the cumulative distribution function of a standard normal $\Phi(x)=\Pr(X \le x)$.

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Hint

$$\begin{align*} P(\text{at least}\,\, a\,\,\text{and}\,\, \text{at most}\,\, b) & = P(a\leq X \leq b)\\ &= P(X \leq b)-P(X\lt a). \end{align*} $$

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In your formula $1 - P(max\space x)$, 1 represents $P(max = "theoretical\space maximum")$. If your max has to be some $M < "theoretical\space maximum"$, it becomes: $P(max\space M)-P(max \space x)$.

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