SOLUTION: The sample space of a random experiment is {a, b, c, d, e, f}, and each outcome is equally likely. A random variable is defined as follows Outcome a b c d e f X 0 0 1.5 1.5 2 3

Question 1192779: The sample space of a random experiment is {a, b, c, d, e, f}, and each outcome is equally likely. A random variable is defined as follows
Outcome a b c d e f
X 0 0 1.5 1.5 2 3
Determine the probability mass function of X.

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
The **probability mass function (PMF)** of a random variable \( X \) provides the probability of each possible value of \( X \). Here's how we determine the PMF:
### Step 1: Analyze the sample space and probabilities
The sample space is \( S = \{a, b, c, d, e, f\} \), and each outcome is equally likely. Since there are 6 outcomes:
\[
P(\text{each outcome}) = \frac{1}{6}.
\]
### Step 2: Group outcomes by the values of \( X \)
The random variable \( X \) assigns values to each outcome as follows:
- \( X(a) = 0 \),
- \( X(b) = 0 \),
- \( X(c) = 1.5 \),
- \( X(d) = 1.5 \),
- \( X(e) = 2 \),
- \( X(f) = 3 \).
We can group the outcomes by the values of \( X \):
- \( X = 0 \): Outcomes \( a, b \) → Probability = \( P(X = 0) = P(a) + P(b) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \).
- \( X = 1.5 \): Outcomes \( c, d \) → Probability = \( P(X = 1.5) = P(c) + P(d) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \).
- \( X = 2 \): Outcome \( e \) → Probability = \( P(X = 2) = P(e) = \frac{1}{6} \).
- \( X = 3 \): Outcome \( f \) → Probability = \( P(X = 3) = P(f) = \frac{1}{6} \).
### Step 3: Write the PMF
The PMF of \( X \) is:
\[
P(X = x) =
\begin{cases}
\frac{1}{3} & \text{if } x = 0, \\
\frac{1}{3} & \text{if } x = 1.5, \\
\frac{1}{6} & \text{if } x = 2, \\
\frac{1}{6} & \text{if } x = 3, \\
0 & \text{otherwise.}
\end{cases}
\]
Answer by ikleyn(52788) (Show Source): You can put this solution on YOUR website!
.

This problem does not determine and does not prescribe any connection between
outcomes a, b, c, d, e, f and values of the random variable X.

What the problem asks, is to determine possible values of P(0), P(1.5), P(2) and P(3).

They can be any four arbitrary real values between 0 and 1;
the only restrictions is that their sum must be 1: P(0) + P(1.5) + P(2)+ P(3) = 1.

But the problem does not prescribe any relations between the outcomes a, b, c, d, d, f
and values of P.

So, in my view, the problem makes no sense, since its different parts
are not connected neither logically, nor physically one with another.

The post by @CPhill produces some semblance of a solution,
but how the problem itself is worded and posed - is NONSENSE.

My general impression is that in this post
two different problems are mistakenly mixed in one.

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