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

 July 21st, 2019, 05:44 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, 02:55 PM #2 Global Moderator   Joined: May 2007 Posts: 6,855 Thanks: 744 $X\ne Y$, 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. July 23rd, 2019, 10:31 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 10:35 AM. July 23rd, 2019, 11:25 AM   #4
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Quote:
 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. Tags ≠x, ≠z, probability, statistics, values, x≠y Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bwilliams632 Calculus 9 March 9th, 2019 12:26 PM Jaket1 Probability and Statistics 2 October 21st, 2017 02:25 PM Mndicedood Probability and Statistics 4 April 4th, 2014 11:25 AM space55 Calculus 0 October 10th, 2010 04:23 PM zolyx Algebra 3 October 28th, 2009 03:57 PM

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