The Joint Probability function of two discrete random variable X and Y is given by
f(x,y) = c(2x+y), for x= 0,1,2 and Y= 0,1,2,3. Find Joint probability distribution and
also evaluate following
a) Constant C b) P(x=2, y=1) c) P[(x>=1), y<=2]
To find the joint probability distribution, we need to determine the value of the constant c and calculate the probability for each possible combination of values for X and Y.
a) To find the constant c, we need to normalize the joint probability function by ensuring that the sum of all probabilities is equal to 1.
Summing up all the possible values of f(x,y) can help us calculate c.
c = 1 / (Sum of all possible values of f(x,y))
In this case, we have:
f(x,y) = c(2x + y)
Summing up all the possible values of f(x,y):
c [(2(0) + 0) + (2(1) + 0) + (2(2) + 0)] + c [(2(0) + 1) + (2(1) + 1) + (2(2) + 1) + (2(3) + 1)]
= c(0 + 2 + 4 + 0 + 1 + 3 + 5 + 7 + 9 + 11)
= c(42)
Since the sum of all possible values of f(x,y) should be equal to 1, we have:
1 = c(42)
Solving for c, we get:
c = 1 / 42
Therefore, the constant c is equal to 1/42.
b) P(x = 2, y = 1) refers to the probability that X takes the value 2 and Y takes the value 1.
To calculate this probability, we substitute the values of x and y into the joint probability function and multiply it by the constant c.
P(x = 2, y = 1) = f(2, 1) * c
Plugging in the values:
P(x = 2, y = 1) = c(2(2) + 1)
P(x = 2, y = 1) = (1/42)(4 + 1)
P(x = 2, y = 1) = 5/42
Therefore, the probability P(x = 2, y = 1) is 5/42.
c) P[(x >= 1), y <= 2] refers to the probability that X is greater than or equal to 1, and Y is less than or equal to 2.
We need to sum up the probabilities for all combinations of x and y that satisfy this condition.
P[(x >= 1), y <= 2] = P(x = 1, y = 0) + P(x = 1, y = 1) + P(x = 1, y = 2) + P(x = 2, y = 0) + P(x = 2, y = 1) + P(x = 2, y = 2)
Plugging in the values and using the constant c:
P[(x >= 1), y <= 2] = [(1/42)(2(1) + 0)] + [(1/42)(2(1) + 1)] + [(1/42)(2(1) + 2)] + [(1/42)(2(2) + 0)] + [(1/42)(2(2) + 1)] + [(1/42)(2(2) + 2)]
P[(x >= 1), y <= 2] = (1/42)(2 + 0) + (1/42)(2 + 1) + (1/42)(2 + 2) + (1/42)(4 + 0) + (1/42)(4 + 1) + (1/42)(4 + 2)
P[(x >= 1), y <= 2] = (1/42)(2 + 2 + 4 + 4 + 5 + 6)
P[(x >= 1), y <= 2] = (1/42)(23)
P[(x >= 1), y <= 2] = 23/42
Therefore, the probability P[(x >= 1), y <= 2] is 23/42.