Question 5

Mathematics Statistics Easy

If X denotes the number obtained on the uppermost face of cubit die when it is tossed, then \(E(X)\) is:

(A) \(2/7\)
(B) \(7/2\)
(C) 1
(D) \(1/2\)
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

The term 'cubit die' is understood to mean a standard six-sided cubic die. Let \(X\) be the random variable representing the number on the uppermost face after a toss.

1. Define the possible outcomes and their probabilities:
The set of all possible outcomes (the sample space) for a single die roll is \(S = \{1, 2, 3, 4, 5, 6\}\).
For a fair die, each of these six outcomes has an equal probability of occurring:
\[P(X=x) = \frac{1}{6} \quad \text{for any } x \in S\]
2. Use the formula for Expected Value (E(X)):
The expected value of a discrete random variable is the sum of each possible outcome multiplied by its corresponding probability.
\[E(X) = \sum_{i=1}^{n} x_i \cdot P(x_i)\]
3. Calculate the Expected Value:
We apply the formula by substituting the outcomes and their probabilities:
\[\begin{aligned} E(X) &= (1 \times \frac{1}{6}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{6}) + (4 \times \frac{1}{6}) + (5 \times \frac{1}{6}) + (6 \times \frac{1}{6}) \\ &= \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) \\ &= \frac{1}{6} (21) \\ &= \frac{21}{6} \end{aligned}\]
Simplifying the final fraction gives:
\[E(X) = \frac{7}{2}\]
Result:
The expected value \(E(X)\) of a single roll of a fair six-sided die is \(\frac{7}{2}\).

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