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There are $n$ normally distributed, independent random variables $X_1, \ldots, X_n$ with distinct means $\mu_1, \ldots \mu_n$ and standard deviations $\sigma_1, \ldots, \sigma_n$. Say, I have $n$ outcomes -- one from each random variable: $x_1, \ldots, x_n$. I would like to compute the CDF value for these outcomes all together with permutations (i.e. the order of the outcomes $x_1, \ldots, x_n$ does not matter). Say, for $n=3$ I would like to compute the area of the following union:

$$ \{X_1 \leq x_1, X_2 \leq x_2 , X_3 \leq x_3\} \cup \\ \{X_1 \leq x_1, X_2 \leq x_3 , X_3 \leq x_2\} \cup \\ \{X_1 \leq x_2, X_2 \leq x_1 , X_3 \leq x_3\} \cup \\ \{X_1 \leq x_3, X_2 \leq x_1 , X_3 \leq x_2\} \cup \\ \{X_1 \leq x_2, X_2 \leq x_3 , X_3 \leq x_1\} \cup \\ \{X_1 \leq x_3, X_2 \leq x_2 , X_3 \leq x_1\} $$

As far as I understand it is not equal to the product of $P(X_1 \leq x_1)\times, \ldots, \times P(X_n \leq x_n)$.

For example, consider a discrete distribution. Say, we roll two dices. What is the probability that we have one dice $\geq4$ and the other dice $\geq2$? So, here we have two discrete random variables -- each corresponding to one dice: $Y_1$ and $Y_2$. We want to compute the probability of the following set: $$ \{Y_1 \geq 4, Y_2 \geq 2\} \cup \{Y_1 \geq 2, Y_2 \geq 4\} $$

I know that the answer is 21/36 or 7/12 (I considered all the 36 combinations on the paper). On the image below the red cells correspond to the combinations that do not satisfy both criteria. In total, there are 21 green cells out of the 36 cells in total, thus the answer should be 21/36:

enter image description here

I can get the value using the following formula: $$ \begin{align} P( set ) & = P( \{Y_1 \geq 4, Y_2 \geq 2\} \cup \{Y_1 \geq 2, Y_2 \geq 4\} ) \\ & = P(Y_1 \geq 4, Y_2 \geq 2) + P(Y_1 \geq 2, Y_2 \geq 4) - P(Y_1 \geq 4, Y_2 \geq 4) \\ & = P(Y_1 \geq 4)P(Y_2 \geq 2) + P(Y_1 \geq 2)P(Y_2 \geq 4) - P(Y_1 \geq 4)P(Y_2 \geq 4) \\ & = \frac{3}{6}\times\frac{5}{6} + \frac{5}{6}\times\frac{3}{6} - \frac{3}{6}\times\frac{3}{6} = \frac{21}{36} \end{align} $$

However, I can't figure out the formula to compute such probability for any number of dice rolling $n$ with a given restriction values $y_1, \ldots, y_n$. I will be grateful for any help or references to the literature. Thanks!

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    $3\times5=15$ and not $20$. And what is this $21$? The correct answer is $\frac{15}{36}$, the product...2017-01-27
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    I repeat my question: Why $21$?2017-01-27
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    Hello, zoli! Thank you for the remarks -- I have updated my question :)2017-01-27
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    You count twice certain outcomes. Se the answer below:2017-01-27

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