In Problem 2, suppose that the white balls are numbered, and...

In Problem 2, suppose that the white balls are numbered, and let $Y_{i}$ equal 1 if the $i$ th white ball is selected and 0 otherwise. Find the joint probability mass function of
(a) $Y_{1}, Y_{2}$
(b) $Y_{1}, Y_{2}, Y_{3}$.

Key Concepts

Joint Probability Mass Function (PMF)

The joint probability mass function gives the probability that each of a set of discrete random variables takes on a specific value simultaneously. It provides a complete description of the relationship between multiple random variables and is essential for analyzing their combined behavior in multivariate probability settings.

Combinatorial Probability

Combinatorial probability involves using counting techniques, such as combinations and permutations, to determine the likelihood of various outcomes in a finite sample space. This approach is key in problems involving selections without replacement, as it helps in calculating the probabilities of specific arrangements or selections among items.

Indicator Random Variables

Indicator random variables take on the value 1 if a certain event occurs and 0 otherwise. They simplify probability problems by reducing events to binary outcomes and are particularly useful when counting the number of times an event occurs in a series of experiments.

Transcript

00:01 Hi guys, in this problem from the given information we observe that three poles are chosen without replacement from an urn consisting of five white and eight red balls.

00:19 So supposing that the white bowl are numbered and consider y is the random variable that defines as yi is equal to 1.

00:33 And 0 1 f the ice white ball is selected and 0 otherwise okay then we have part a we need to calculate with a probability mass function y 1 y2 so the probability of selecting first and second white balls is chosen is y 1 is y 1 is equal to 1 and y2 is equal to 1 so there exists 13 c3 ways to choose three poles out of the 13 of them so if the first the two poles have been choosing there exist 10 sorry 11 c1 ways to do that so probability of y1 is equal to 1 and y2 is equal to 1 so there is 11 c3 over 13 c3 okay so this is 11 over 286 which is 1 over 26 okay okay then we need to find the probability of y1 is equal to 1 and y 2 is equal to 0 so this is 11c2 over 13 c3.

02:13 Okay, so this is 5 over 26.

02:19 Then probability of y1 is equal to 0, y2 is equal to 1.

02:28 So this is 11c2 over 13, c3 so this is 5 over 26 okay now the probability to choose both poles are are not white such as that probability of y1 is 0 and y2 is 0 so this is 11 c3 over 13 c3 okay so so this is 15 over 26.

03:14 Okay, in part b is the number of variables is 3.

03:17 So there are 3 squared which is 10 ways.

03:21 So probability of y1 is 1, y 2 is 1, and y 3 is 1.

03:32 This is 10 c 0 over 13 c3...

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