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Sagot :
The joint distribution of the numbers of the three colours in the sample without replacement is:
[tex]\mathbf{P(A=a,B=b,C=c) = \dfrac{\Big( ^{p}_{ A} \Big) \Big( ^{q}_{ B} \Big) \Big( ^{r}_{ C} \Big) }{ \mathbf{\Big( ^{p+q+r}_{ \ \ \ n} \Big) } }\ \ \ \ where; n = A+B+C}[/tex]
Let consider A, B, and C to denote the three variables that are black, white, and red balls in the sample.
i.e.
- n = A + B + C
Now, the numbers of ways 'n' balls are chosen without replacement in an urn that comprises of p black balls, q white balls, and r red balls can be computed as follows:
[tex]\mathbf{\implies \Big( ^{p+q+r}_{ \ \ \ n} \Big) }[/tex]
Now, the number of ways whereby A black balls can be chosen from p black balls is expressed as:
[tex]\mathbf{\implies \Big( ^{p}_{ A} \Big) }[/tex]
The number of ways whereby B white balls can be chosen from q white balls is expressed as:
[tex]\mathbf{\implies \Big( ^{q}_{ B} \Big) }[/tex]
The number of ways whereby C red balls can be chosen from r red balls is expressed as:
[tex]\mathbf{\implies \Big( ^{r}_{ C} \Big) }[/tex]
Therefore, we can conclude that the joint distribution of the numbers of the three colours in the sample without replacement is:
[tex]\mathbf{P(A=a,B=b,C=c) = \dfrac{\Big( ^{p}_{ A} \Big) \Big( ^{q}_{ B} \Big) \Big( ^{r}_{ C} \Big) }{ \mathbf{\Big( ^{p+q+r}_{ \ \ \ n} \Big) } }\ \ \ \ where; n = A+B+C}[/tex]
Learn more about joint distribution here:
https://brainly.com/question/17283589?referrer=searchResults
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