My Math Forum What is the probability that X, Y and Z take different values (X≠Y, Y ≠Z, Z ≠X)

 July 21st, 2019, 04:44 AM #1 Newbie   Joined: Jun 2019 From: London Posts: 13 Thanks: 0 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?
 July 21st, 2019, 01:55 PM #2 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 717 $X\ne Y$, 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 01:24 AM.
 July 23rd, 2019, 09:31 AM #3 Newbie   Joined: Jun 2019 From: London Posts: 13 Thanks: 0 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 09:35 AM.
July 23rd, 2019, 10:25 AM   #4
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 Originally Posted by mathodman25 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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