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An urn contains p black balls, q white balls, and r red balls; and n balls are chosen without replacement. a. Find the joint distribution of the numbers of black, white, and red balls in the sample. b. Find the joint distribution

Sagot :

ajeigbeibraheem

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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