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July 21st, 2019, 05:44 AM   #1
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What is the probability that X, Y and Z take different values (X≠Y, Y ≠Z, Z ≠X)

Independent random variables X, Y and Z take only integer values: X - from 1 to 15 with probability 1/15, Y - from 1 to 10 with probability 1/10, Z - from 1 to 8 with probability 1/8. What is the probability that X, Y and Z take different values (X≠Y, Y ≠Z, Z ≠X)? I have an idea of writing down all possible combinations, but i want to know is there some easier solution?
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July 21st, 2019, 02:55 PM   #2
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$X\ne Y$, prob.$=\frac{1}{3}+\frac{2}{3}\times \frac{9}{10}$.
$Y\ne Z$, prob.$=\frac{1}{5}+\frac{4}{5}\times \frac{7}{8}$.
$X\ne Z$, prob.$=\frac{7}{15}+\frac{8}{15}\times \frac{7}{8}$.
I assume you can figure out what it means.

Last edited by mathman; July 22nd, 2019 at 02:24 AM.
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July 23rd, 2019, 10:31 AM   #3
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And now multiply all three? But how did you get these probabilities? Could you please explain a little bit more detailed how did you get the first one for example?

Last edited by mathodman25; July 23rd, 2019 at 10:35 AM.
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July 23rd, 2019, 11:25 AM   #4
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Quote:
Originally Posted by mathodman25 View Post
And now multiply all three? But how did you get these probabilities? Could you please explain a little bit more detailed how did you get the first one for example?
$X \in \{11, 12, 13, 14, 15\} \text{ with probability }\dfrac 1 3$

$\text{If $X \not \in \{11, 12, 13, 14, 15\}$ then the probability that $X$ and $Y$ are chosen differently is $\dfrac{10}{15} \cdot \dfrac{9}{10} = \dfrac 2 3 \cdot \dfrac{9}{10}$}$

$\text{This is because there are $10$ possible values of $X$ and once we've chosen $X$ only $9$ possible values of $Y$}$

$\text{We choose one of the $10~ X$ values with probability $\dfrac{10}{15}$ and then the $Y$ value with probability $\dfrac{9}{10}$}$

$\text{So the total probability is the sum of the probabilities of these two cases}$

$P=\dfrac 1 3 + \dfrac 2 3 \dfrac{9}{10} \text{ as mathman posted}$

Now see if you can derive the others.
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