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Let X and Y be discrete random variables with joint probability mass function; P(X = x,Y = y) = 2x + y/12 , for (x,y) = (0,1),(0,2),(1,2),(1,3) 0, elsewhere Determine the marginal probability function of X and show that it is a probability function.

David Gbadam
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  • What have you tried? – David G. Stork Apr 30 '20 at 20:28
  • This is very similar to your other problem. Make your table. Verify that it is a probability function by showing that the entries in the table all add up to a total of $1$. Find your marginal probability function by adding subtotals to columns or adding subtotals to rows (depending on which way you oriented your table). – JMoravitz Apr 30 '20 at 21:03
  • Also... you should be more careful with how you type things 2x+y/12 reads like $2x + (\frac{y}{12})$, not as $\frac{(2x+y)}{12}$ – JMoravitz Apr 30 '20 at 21:04
  • I have been able to fix them into a table but after I drew the table, I did not get exactly 1. For both problems when I add their marginal totals. I got figures close to 1 instead but not 1 itself – David Gbadam Apr 30 '20 at 21:29
  • "I did not get exactly 1" Show your work and show us what you did get. Note, $\frac{1}{12}\approx 0.083$ but only approximately. It can be helpful to keep things as fractions. If your error is like how someone might think that $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$ is not equal to $1$ simply because $0.33+0.33+0.33=0.99\neq 1$... then you really need to stop whatever bad habit is causing that. – JMoravitz Apr 30 '20 at 21:56
  • I kept them as fractions and had 49/48 which is equivalent to 1.020... – David Gbadam Apr 30 '20 at 22:12
  • Then you have an arithmetic error somewhere. $((2\cdot 0 + 1) + (2\cdot 0 + 2)+(2\cdot 1+2)+(2\cdot 1+3))/12 = (1+2+4+5)/12=12/12=1$. I have no idea why you got a denominator of $48$ when every fraction involved had a denominator of twelve... – JMoravitz May 01 '20 at 02:52

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