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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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390  Quote:
$\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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≠x, ≠z, probability, statistics, values, x≠y 
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