The joint probability function of X and Y is given by
𝑓(𝑥) = 𝐶(𝑥^2+y^2) ,𝑓𝑜𝑟 𝑥 = −1, 0, 1, 3 𝑎𝑛𝑑 𝑦 = −1, 2, 3. Then find
a. The value of constant C
b. 𝑃(𝑥 ≤ 1, 𝑦 > 2)
c. 𝑃(𝑥 + 𝑦 > 2)
d. 𝑃(𝑥 = 0, 𝑦 ≤ 2)
To find the value of constant C, we can use the fact that the joint probability function must sum to 1 over all possible values of x and y. So we have:
1 = ΣΣ𝐶(𝑥^2+𝑦^2)
Let's compute the sum:
1 = 𝐶((-1)^2+(-1)^2) + 𝐶((-1)^2+2^2) + 𝐶((-1)^2+3^2) + 𝐶(0^2+(-1)^2) + 𝐶(0^2+2^2) + 𝐶(0^2+3^2) + 𝐶(1^2+(-1)^2) + 𝐶(1^2+2^2) + 𝐶(1^2+3^2) + 𝐶(3^2+(-1)^2) + 𝐶(3^2+2^2) + 𝐶(3^2+3^2)
Simplifying the expression:
1 = 2𝐶 + 5𝐶 + 10𝐶 + 𝐶 + 4𝐶 + 9𝐶 + 2𝐶 + 5𝐶 + 10𝐶 + 10𝐶 + 13𝐶 + 18𝐶
Combining like terms:
1 = 91𝐶
Dividing both sides by 91:
𝐶 = 1/91
a. The value of constant C is 1/91.
To find 𝑃(𝑥 ≤ 1, 𝑦 > 2), we need to sum up the probabilities of all the x and y values that satisfy this condition. That is, we want to find:
𝑃(𝑥 ≤ 1, 𝑦 > 2) = 𝑃(𝑥 = -1, 𝑦 > 2) + 𝑃(𝑥 = 0, 𝑦 > 2) + 𝑃(𝑥 = 1, 𝑦 > 2)
Using the joint probability function f(x), we have:
𝑃(𝑥 ≤ 1, 𝑦 > 2) = 𝐶((-1)^2+2^2) + 𝐶(0^2+2^2) + 𝐶(1^2+2^2)
Substituting the value of C:
𝑃(𝑥 ≤ 1, 𝑦 > 2) = (1/91)(5+4+5)
Simplifying the expression:
𝑃(𝑥 ≤ 1, 𝑦 > 2) = 14/91
b. 𝑃(𝑥 ≤ 1, 𝑦 > 2) is 14/91.
To find 𝑃(𝑥 + 𝑦 > 2), we need to sum up the probabilities of all the x and y values that satisfy this condition. That is, we want to find:
𝑃(𝑥 + 𝑦 > 2) = 𝑃(𝑥 = -1, 𝑦 > 2) + 𝑃(𝑥 = 0, 𝑦 > 2) + 𝑃(𝑥 = 1, 𝑦 > 2) + 𝑃(𝑥 = 3, 𝑦 > 2)
Using the joint probability function f(x), we have:
𝑃(𝑥 + 𝑦 > 2) = 𝐶((-1)^2+3^2) + 𝐶(0^2+3^2) + 𝐶(1^2+3^2) + 𝐶(3^2+3^2)
Substituting the value of C:
𝑃(𝑥 + 𝑦 > 2) = (1/91)(10+9+10+18)
Simplifying the expression:
𝑃(𝑥 + 𝑦 > 2) = 47/91
c. 𝑃(𝑥 + 𝑦 > 2) is 47/91.
To find 𝑃(𝑥 = 0, 𝑦 ≤ 2), we can use the joint probability function f(x). We have:
𝑃(𝑥 = 0, 𝑦 ≤ 2) = 𝐶(0^2+(-1)^2) + 𝐶(0^2+2^2) + 𝐶(0^2+3^2)
Substituting the value of C:
𝑃(𝑥 = 0, 𝑦 ≤ 2) = (1/91)(1+4+9)
Simplifying the expression:
𝑃(𝑥 = 0, 𝑦 ≤ 2) = 14/91
d. 𝑃(𝑥 = 0, 𝑦 ≤ 2) is 14/91.