Question

Let an unbiased die be thrown and the random variable X be the number appears on its top. Then the expectation of X is

• A. 1
• B. \(\frac{1}{2}\)
• C. \(\frac{7}{2}\)
• D. \(\frac{6}{2}\)

Hint:

For any process with equally likely integer outcomes from 1 to n (like a fair die or a spinner), the expected value is simply the average of the first and last outcome: \(\frac{1 + n}{2}\). For a die, this is \(\frac{1 + 6}{2} = \frac{7}{2} = 3.5\).

Answer 1

The Correct Option is C

Step 1: Definition:
The expected value of a discrete random variable is the probability-weighted average of its possible values. For a fair die, all six faces have equal probability.
Step 2: Formula:
The expected value \(E[X]\) of a discrete random variable X is: \[ E[X] = \sum_{i} x_i P(X = x_i) \]where \(x_i\) are the possible values of X and \(P(X = x_i)\) is the probability of each value.
Step 3: Calculation:
For a single throw of a fair die:
Possible outcomes (values of X) are \(\{1, 2, 3, 4, 5, 6\}\).
Each outcome has the same probability: \[ P(X=1) = P(X=2) = P(X=3) = P(X=4) = P(X=5) = P(X=6) = \frac{1}{6} \]Applying the formula: \[ E[X] = \left(1 \times \frac{1}{6}\right) + \left(2 \times \frac{1}{6}\right) + \left(3 \times \frac{1}{6}\right) + \left(4 \times \frac{1}{6}\right) + \left(5 \times \frac{1}{6}\right) + \left(6 \times \frac{1}{6}\right) \]Factor out \(\frac{1}{6}\): \[ E[X] = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) \]The sum is \(1+2+3+4+5+6 = 21\). \[ E[X] = \frac{1}{6} (21) = \frac{21}{6} \]Simplify: \[ E[X] = \frac{7}{2} \] Step 4: Result:
The expected value of X is \(\frac{7}{2}\) or 3.5, which is option (3).