Joint distribution of a discrete random variable

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SUMMARY

The discussion focuses on calculating the joint distribution of two random variables, X and Y, derived from drawing 4 balls without replacement from an urn containing 6 balls numbered 1 to 6. X represents the minimal number drawn, while Y represents the maximal number drawn. Participants suggest enumerating all possible combinations, noting that there are 15 combinations in total, calculated using the binomial coefficient $\binom{6}{4}$. This approach simplifies the probability calculations for specific pairs of values for X and Y.

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  • Understanding of discrete random variables
  • Familiarity with joint distributions
  • Knowledge of combinatorial mathematics, specifically binomial coefficients
  • Basic probability theory
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  • Study the concept of joint probability distributions in depth
  • Learn how to calculate binomial coefficients and their applications
  • Explore enumeration techniques for probability problems
  • Investigate the properties of minimal and maximal values in sampling without replacement
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Yankel
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Hello all

I have this question I am trying to solve.

In an urn there are 6 balls, numbered: 1,2,3,4,5,6. We take 4 balls outs, without replacement.

X - the minimal number we see
Y - the maximal number we see

I need to joint distribution.

I understand that X is getting the values 1,2,3 while Y 4,5,6.

The problem is calculating the probabilities. How do I calculate the probability that X = 1 and Y = 5 ? What about the probability that X = 2 and Y = 6 ? And so on...

thanks !
 
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Yankel said:
Hello all

I have this question I am trying to solve.

In an urn there are 6 balls, numbered: 1,2,3,4,5,6. We take 4 balls outs, without replacement.

X - the minimal number we see
Y - the maximal number we see

I need to joint distribution.

I understand that X is getting the values 1,2,3 while Y 4,5,6.

The problem is calculating the probabilities. How do I calculate the probability that X = 1 and Y = 5 ? What about the probability that X = 2 and Y = 6 ? And so on...

thanks !

Hi Yankel,

How about enumerating them all?
There are only $\binom 6 4 = 15$ possible combinations.
 
A little bit hard to count them all, but it works ! Thanks
 

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