The joint probability mass function of two discrete random...

The joint probability mass function of two discrete random variables X and Y is given by | Y X | 0 | 1 | 2 | |---|---|---|---| | 0 | 1/6 | 1/12 | 1/12 | | 1 | 0 | 1/6 | 1/3 | | 2 | 0 | 0 | 1/6 | (a) Find the conditional probability mass function of Y given X = 1. (b) Find the conditional probability mass function of X given Y = 1. (c) Determine whether X and Y are independent.

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00:01 Hi, i'm david and i'm here to help you answer your question.

00:04 Now let me bring up your question here.

00:06 In the question here, we are given the john probability mass function under two discrete random variable x and y.

00:13 In the question a, once you find the conditional probability mass function of the y given x equal to 1.

00:20 So we will make the table for the y here, given that the x is equal to 1.

00:29 So the y will take the value of the 0.

00:31 1 and 2 and there could be the probability of the y even x equal to 1 and now we will have end the x equal to 1 it means that we only focus on the the middle column here and then we will see that we have to add the total probability for these three of them so we have the total it will equal to 1 over 12 plus 1 of over 6 and then we get equal to 1 over 4.

01:05 So therefore the probability for the y equal to 0, we will turn the value of the 1 over 12 dividing by the 1 over 4.

01:16 And then if we do it, we get equal to the 1 over 12 divided the answer equal to the 1 over 3.

01:27 And for the volume the 1, we will take the 1 over 6 divided by the 1 over 4.

01:33 So we get equal to the 2 out of 3.

01:36 And the last one we have the 0 there.

01:39 0 divided by 1 over 4 and equal to the 0.

01:43 So therefore this will be the conditional probability mass function on the y given x equal to 1.

01:52 Now for the question b, once you find the conditional probability mass function under x given then the y equal to 1...

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