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: Understanding the Concept:
The expectation (or expected value) of a discrete random variable is the probability-weighted average of all its possible values. For an unbiased die, each of the six faces has an equal probability of appearing.
Step 2: Key Formula or Approach:
The formula for 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 occurring.
Step 3: Detailed Explanation:
For a single throw of an unbiased die:
The set of possible outcomes (values of X) is \(\{1, 2, 3, 4, 5, 6\}\).
Since the die is unbiased, the probability of each outcome is the same: \[ P(X=1) = P(X=2) = P(X=3) = P(X=4) = P(X=5) = P(X=6) = \frac{1}{6} \] Now, we apply the expectation 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 the common probability \(\frac{1}{6}\): \[ E[X] = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) \] The sum of the numbers is \(1+2+3+4+5+6 = 21\). \[ E[X] = \frac{1}{6} (21) = \frac{21}{6} \] Simplify the fraction by dividing the numerator and denominator by 3: \[ E[X] = \frac{7}{2} \] Step 4: Final Answer:
The expectation of X is \(\frac{7}{2}\) or 3.5. This matches option (3).