Question

Suppose that 3 balls are chosen without replacement from an urn consisting of 5 white and 8 red balls. Let X i equal 1 if the i th ball selected is white, and let it equal 0 otherwise. Give the joint probability mass function of X 1 , X 2 ; X 1 , X 2 , X 3 .

Answer 1

Answer:

Mixed probabilty

Step-by-step explanation:

What is the probability that the first and second ball chosen are both targets, that is, X1 = 1 X2 = 2?

For this we must study the simple probability of each, for example for the target in that event is 5/13. However for the second chosen but also white in the second event where there is already a total of 12 balls is 4/12.

Thus

$P (X1 = 1, X2 = 1) = \frac{5}{13} * \frac{4}{12} = \frac{5}{39}$

In this way is generated for all probabilities

$P (X1 = 1, X2 = 0) = \frac{5}{13} *\frac{8}{12} = \frac{10}{39}\\P (X1 = 0, X2 = 1) = \frac{8}{13} * \frac{5}{12} = \frac{10}{39}\\P (X1 = 0, X2 = 0) = \frac{8}{13} *\frac{7}{12} =\frac{4}{39}$

b) In the same way analogous to the past example we can perform all cases of combinations, therefore it would be so

$P (X1 = 1, X2 = 1) = \frac{5}{13}*\frac{4}{12}*\frac{3}{11}= \frac{5}{143}$

$P (X1 = 0, X2 = 1) = \frac{8}{13} *\frac{5}{12} * \frac{4}{11} = \frac{40}{429}$

$P (X1 = 1, X2 = 1) = \frac{5}{13}*\frac{8}{12}*\frac{4}{11} = \frac{40}{429}$

$P (X1 = 1, X2 = 0) =\frac{5}{13}*\frac{4}{12}*\frac{8}{11} = \frac{40}{429}$

$P(X1 = 0, X2 = 1) = \frac{8}{13}* \frac{7}{12}*\frac{5}{11}=\frac{70}{429}$

$P (X1 = 0, X2 = 0) = \frac{8}{13} * \frac{7}{12} * \frac{5}{11} = \frac{70}{429}$

$P (X1 = 1, X2 = 0) = \frac{8}{13} * \frac{7}{12} * \frac{5}{11} = \frac{70}{429}$

$P (X1 = 0, X2 = 0) = \frac{8}{13} *\frac{7}{12} * \frac{6}{11} = \frac{28}{143}$