Answer:
The function P(X=x) = c(1 - (7/2)^x) for x = 1, 2, 3 is not a valid probability mass function for any value of c, as it violates both conditions.
Explanation:
To determine if the function P(X=x) = c(1 - (7/2)^x) for x = 1, 2, 3 is a valid probability mass function for any c, we need to verify if it satisfies the properties of a probability mass function.
A valid probability mass function must satisfy two conditions:
The probability for each value of x must be between 0 and 1.
The sum of the probabilities for all possible values of x must equal 1.
Let's evaluate these conditions for the given function:
The probability for each value of x must be between 0 and 1:
For x = 1, 2, 3, the function becomes:
P(X=1) = c(1 - (7/2)^1)
= c(1 - 7/2)
= c(-5/2)
P(X=2) = c(1 - (7/2)^2)
= c(1 - 49/4)
= c(-45/4)
P(X=3) = c(1 - (7/2)^3)
= c(1 - 343/8)
= c(-335/8)
Since c is a constant, the function can be negative for certain values of c.
For example, if c is negative, the probabilities P(X=x) will be negative as well.
This violates the condition that probabilities must be between 0 and 1.
Therefore, the given function is not valid for any value of c.
The sum of the probabilities for all possible values of x must equal 1: Let's calculate the sum of the probabilities:
P(X=1) + P(X=2) + P(X=3)
= c(-5/2) + c(-45/4) + c(-335/8)
= -5c/2 - 45c/4 - 335c/8
= -20c/8 - 90c/8 - 335c/8
= (-20c - 90c - 335c)/8
= (-445c)/8
Since the sum of the probabilities is dependent on c, and c can take any value, it is not possible to guarantee that this sum will equal 1.
Therefore, the given function does not satisfy the condition of the sum of probabilities equaling 1.
Thus,
The function P(X=x) = c(1 - (7/2)^x) for x = 1, 2, 3 is not a valid probability mass function for any value of c, as it violates both conditions.