Solve the following If the join probability distribution of X and Y iss given by f(x,y)=x+y30 f or x=0, 2, 3: y=0, 1, 2 Find a) P(X≤2, Y=1). b) P(X&gt;2, Y≤1). c) P(X&gt;Y).

Expert answers to "Solve the following If the join probability distribution..."

UkusakazaL

Answered question

2021-08-04

Solve the following
If the join probability distribution of X and Y iss given by

f (x, y) = \frac{x + y}{30}

f or

x = 0, 2, 3 :

y = 0, 1, 2

Find
a)

P (X \leq 2, Y = 1) .

P (X > 2, Y \leq 1) .

P (X > Y) .

Answer & Explanation

Ayaana Buck

Beginner2021-08-04Added 2 answers

Step 1
Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.
Here,
$\begin{array}{cc} x & y & f (x, y) \\ 0 & 0 & 0 \\ 2 & 0 & 0.066667 \\ 3 & 0 & 0.1 \\ 0 & 1 & 0.033333 \\ 2 & 1 & 0.1 \\ 3 & 1 & 0.133333 \\ 0 & 2 & 0.066667 \\ 2 & 2 & 0.133333 \\ 3 & 2 & 0.166667 \end{array}$
Step 2
a) $P (X \leq 2, Y = 1) = P (X = 0, Y = 1) + P (X = 2, Y = 1)$
$= f (0, 1) + f (2, 1)$
$= 0.033333 + 0.1$
$= 0.133333$
Step 3
b) $P (X > 2, Y \leq 1) = P (X = 3, Y = 0) + P (X = 3, Y = 1)$
$= f (3, 0) + f (3, 1)$
$= 0.1 + 0.133333$
$= 0.233333$
Step 4
c) $P (X > Y) = P (X - Y > 0)$
$= P (X = 2, Y = 0) + P (X = 3, Y = 0) + P (X = 2, Y = 1) + P (X = 3, Y = 1)$
$+ P (X = 3, Y = 2)$
$= 0.066667 + 0.1 + 0.1 + 0.133333 + 0.166667$
$= 0.566667$

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