Question

What is the mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on 1 face?

• A. 2
• B. 4
• C. 1
• D. 8

Hint:

To calculate the mean of a discrete distribution, think of it as a weighted average. Multiply each possible value by its probability and sum the results. This is a fundamental concept in probability and statistics.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The "mean" of the numbers obtained from a random event like throwing a die refers to the expected value of the outcome. The expected value is a weighted average of all possible outcomes, where each outcome is weighted by its probability.
Step 2: Key Formula or Approach:
The mean or expected value E[X] of a discrete random variable X is calculated as: \[ E[X] = \sum_{i} x_i P(X=x_i) \] where \(x_i\) are the possible outcomes and \(P(X=x_i)\) are their respective probabilities.
Step 3: Detailed Explanation:
First, we need to determine the probability of each outcome. The die has a total of 6 faces.
- The number 1 is on 3 faces, so \(P(X=1) = \frac{3}{6} = \frac{1}{2}\).
- The number 2 is on 2 faces, so \(P(X=2) = \frac{2}{6} = \frac{1}{3}\).
- The number 5 is on 1 face, so \(P(X=5) = \frac{1}{6}\).
(Check: The probabilities sum to 1: \(\frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1\)).
Now, we apply the formula for the expected value: \[ E[X] = (1 \times P(X=1)) + (2 \times P(X=2)) + (5 \times P(X=5)) \] \[ E[X] = \left(1 \times \frac{3}{6}\right) + \left(2 \times \frac{2}{6}\right) + \left(5 \times \frac{1}{6}\right) \] \[ E[X] = \frac{3}{6} + \frac{4}{6} + \frac{5}{6} \] \[ E[X] = \frac{3+4+5}{6} = \frac{12}{6} = 2 \] Step 4: Final Answer:
The mean of the numbers obtained on throwing the die is 2.