Question

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

• A. $5$
• B. $\frac{8}{3}$
• C. $1$
• D. $2$

Hint:

To compute the mean of a die throw quickly, simply sum up the values written on every single face and divide by the total number of faces (6): $\text{Mean} = \frac{1+1+1+2+2+5}{6} = \frac{12}{6} = 2$. This avoids working with fractions entirely!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks for the mathematical mean (expected value) of the score obtained when casting a biased six-sided die whose faces are customized with non-standard values.

Step 2: Key Formula or Approach:
The mean or expected value $\mu = E(X)$ of a discrete random variable is computed using the probability distribution formula:
$$\mu = \sum x_i \cdot P(x_i)$$ Where $x_i$ represents each unique score value on the die and $P(x_i)$ is its corresponding probability of occurrence.

Step 3: Detailed Explanation:
Let's list out the face layout configuration of the six-sided die:
The number 1 is written on 3 faces $\implies P(X = 1) = \frac{3}{6} = \frac{1}{2}$
The number 2 is written on 2 faces $\implies P(X = 2) = \frac{2}{6} = \frac{1}{3}$
The number 5 is written on 1 face $\implies P(X = 5) = \frac{1}{6}$
Now, evaluate the expected value sum by multiplying each outcome by its probability:
$$\mu = \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{3}\right) + \left(5 \times \frac{1}{6}\right)$$ $$\mu = \frac{1}{2} + \frac{2}{3} + \frac{5}{6}$$ Find a common denominator, which is 6:
$$\mu = \frac{3}{6} + \frac{4}{6} + \frac{5}{6}$$ $$\mu = \frac{3 + 4 + 5}{6} = \frac{12}{6} = 2$$ The expected mean value matches option (D).

Step 4: Final Answer:
The mean of the numbers obtained is $2$, which corresponds to option (D).