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 is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
Let X be the number obtained. Each of the six faces is equally likely, so
\( \displaystyle P(X = 1) = \frac{3}{6},\quad P(X = 2) = \frac{2}{6},\quad P(X = 5) = \frac{1}{6} \)
The mean (expected value) is the sum of each value times its probability:
\( \displaystyle E(X) = 1 \times \frac{3}{6} + 2 \times \frac{2}{6} + 5 \times \frac{1}{6} \)
\( \displaystyle = \frac{3 + 4 + 5}{6} = \frac{12}{6} = 2 \)
Why the other options are incorrect
Option (1) 1 — Incorrect. 1 is only the most frequent number on the die (the mode), not the mean.
Option (3) 5 — Incorrect. 5 is the largest number on the die; the mean, 2, is much smaller because 5 is on one face only.
Option (4) \(\displaystyle\frac{8}{3}\) — Incorrect. This is the plain average of the three different numbers, (1 + 2 + 5)/3, which ignores how many faces carry each number.