Question

Let X and Y be discrete random variables with the joint probability function f(x,y) given by the table: х f(x,y) 7 9 11 20.050.15 0.1 Y 30.15 0.40.15 Which of the following is the conditional probability function fxy (x|3) ? X = x 7 9 11 fxry(x3)0.28|0.37.42 X = X 7 9 11 fxıy(x3)0.30.280.42 X = x 9 7 11 fxıy(x|3)|0.2143|0.2143|0.5714 X = X 7 11 fxxy(xl3)|0.21430.57140.2143 ОО None of the above

Answer 1

X and Y are discrete random variables given by the table

	X=7	X=9	X=11	Total
Y=2	0.05	0.15	0.1	0.3
Y=3	0.15	0.4	0.15	0.7
Total	0.2	0.55	0.25	1

We have to find the conditional probability function

fxy(23)

Now, X can take only values 7,9 and 11.

Now,

$f_{X|Y}(x=7|3)$

$=\frac{f(x=7,y=3)}{f(y=3)}$

$=\frac{f(x=7\bigcap y=3)}{f(y=3\bigcap x=7)+f(y=3\bigcap x=9)+f(y=3\bigcap x=11)}$

015 0.15 +0.4 +0.15

$=\frac{0.15}{0.70}$

$=0.2143$

Then,

$f_{X,Y}(x=9|3)$

$=\frac{f(X=9\bigcap y=3)}{f(y=3)}$

$=\frac{f(X=9\bigcap y=3)}{f(y=3\bigcap x=7)+f(y=3\bigcap x=9)+f(y=3\bigcap x=11)}$

$=\frac{0.40}{0.15+0.4+0.15}$

$=\frac{0.40}{0.70}$

$=0.5714$

$f_{X,Y}(x=11|3)$

$=\frac{f(x=11\bigcap y=3)}{f(y=3)}$

$=\frac{f(x=11\bigcap y=3)}{f(y=3\bigcap x=7)+f(y=3\bigcap x=9)+f(y=3\bigcap x=11)}$

015 0.15 +0.4 +0.15

$=\frac{0.15}{0.70}$

$=0.2143$

So, the conditional distribution is

X=x	7	9	11
f(x|3)	0.2143	0.5714	0.2143

The correct answer is option (D).