A discrete random variable’s probabilities must sum to 1. The given values for f(x) (0.49, 0.51, 0.51, 0.51) sum to 2.02, not 1. This means f(x) cannot be the direct probability mass function.
In such cases, it is typically assumed that the given f(x) values are proportional to the actual probabilities, and they need to be normalized.
Here’s how to calculate the probabilities:
1. **Sum the given f(x) values:**
Sum = 0.49 + 0.51 + 0.51 + 0.51 = 2.02
2. **Normalize each f(x) value to find the probabilities P(X=x):**
P(X=x) = f(x) / Sum
* **For X = 1:**
P(X=1) = 0.49 / 2.02 ≈ 0.24257
* **For X = 2:**
P(X=2) = 0.51 / 2.02 ≈ 0.25248
* **For X = 3:**
P(X=3) = 0.51 / 2.02 ≈ 0.25248
* **For X = 4:**
P(X=4) = 0.51 / 2.02 ≈ 0.25248
Therefore, the probabilities of each of the values of X are approximately:
* **P(X=1) ≈ 0.24257**
* **P(X=2) ≈ 0.25248**
* **P(X=3) ≈ 0.25248**
* **P(X=4) ≈ 0.25248**
(Note: The sum of these rounded probabilities is 0.24257 + 0.25248 + 0.25248 + 0.25248 = 1.00001 due to rounding. Using the exact fractions, the sum is precisely 1.)