An urn contains 3 red, 3 black and 4 white balls. Three...

An urn contains 3 red, 3 black and 4 white balls. Three balls are selected at random without replacement from the urn. Let X be the number of red balls and Y be the number of black balls drawn. [3] a. Find the joint probability function $f(x, y)$. Either a formula or a tabulation is acceptable. [2] b. Find the marginal probability function $f_x(x)$ of X. [4] c. Find $\rho = Corr(X, Y)$. [2] d. Why would you expect $\rho$ to have the sign you found?

Transcript

00:01 In this problem, an urn contains 3 black, 2 red and 3 green balls.

00:08 Let x is the number of red balls and y be the number of black balls.

00:15 Right.

00:16 So here x is for red and y is for black balls.

00:27 So here first of all it is asked that if 2 balls are selected from the urn, at random with replacement right then find the joint probability distribution and probability mass function right so okay so this must be like this that a is equals to so you can say that you may have you are having three black balls right and and three black balls are present black and we have red is two red balls right so here what all combinations are possible so we can have p when x equals to zero and y is also equals to zero when no of the balls are black or red you can say so the probability would be three by 8 into 3 by 8 equals to so how is it possible because total number of outcomes would be going to be 8 itself right but here 3 green balls to be there right so that's why the expected outcome is 3 itself so for 1 ball it should be 3 by 8 and other ball it should be 3 by 8 so the total problem becomes 9 by 64 right again px equals to 0 again when y equals to 1 that means 1 is a blackboard right so here this is is equal to 3 by 8 again into 3 by 8 again so this will be equal to 9 by 64 again similarly we are following the same things so these are the probabilities when we are selecting we are having we are just having a sequence like how how are the possibilities possible right so this is y equals to 2 so this will be equal to 1 by i'm sorry 3 by 8 into 3 by 8 that will be again 9 by 64 itself right now for p x equals to 1 and y equals to 0 it should be equals to 2 by 8 into 3 by 8 which is equals to 6 by 64 equals to 3 by 32 now p x x equals to 1 and y equals to 1 equals to 2 by 8 into 3 by 8 again which is equals to 6 by 64 gives us 3 by 32 right this will be p equals to p when x equals to 2 y equals to 0 this will be equal to 2 by 8 into 2 by 8 which is equals to 4 by 64 equals to 1 by 16 right now this is for the part 1 we have seen now for the part 2 part b we can say that if two balls are selected from it at random without replacement then find joint probability distribution and probability mass function right so here f xy if xy equals to right so here it is told that two balls is to be selected right so here the ways should be 2 cx into 3 c y and and it will be 3c 2 minus x minus y here right so this will be upon 8c2 right so this is total number of outcomes so here we can have where x is equal to 0 1 2 and y is equal to 0 1 2 and x is equal y must be less than or equals to 2 right so we will have a we will just write probability mass function which is p when x equals to 0 y equals to 0 then it is equal to 2c 0 into 3c 0 into 3c 2 which is upon 8c2 right this will give us 3 by 28 okay now p when x equals to 0 y equals to 1 then this should be equals to 2c 0 this is a logical thing only that what all balls we can select right so what we are choosing right so this is 3c1 again 3 c1 upon 8c2 this is equal to 9 by 28 right so when p when x equals to 0 and y equals to 2 so it will be giving us 2 c0 multiplied with 3 c2 multiplied with 3 c0 upon 8c 2 which is equal to 3 by 8 8 2 which is equal to 3 by 8 right so now p of x1 x is equal when x equals to 1 and y equals to 0 then this will give us 2c1 into 3c0 into 3c1 upon 8c2 this is equal to 6 by 8 8 equals to 3 by 40 i'm sorry this is 28 right so now p when x is equal to 1 and y is equal to 1 which is equal to 2c1 into 3c1 into 3 c0 upon 8c 2 right so this will be giving us 6 by 28 again which is equal to 3 by 14 right now for p when x is equal to 2 and y equals to 0 it should be equals to 2 c 0 3 c 0 so this will be upon 8c 2 right which is giving us 1 by 28 right so i am writing the table here that how can you it in this way...

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