A spinner is divided into three sectors. The sector marked 1 has an angle of 180°, the sector marked 2 has an angle of 120° and the sector marked 6 has an angle of 60°. The pointer is equally likely to stop at any point of the circle. If X is the number on which the pointer stops, then \(E(X)\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{13}{6}\)
The probability of each number is the angle of its sector divided by 360°:
\( \displaystyle P(X = 1) = \frac{180}{360} = \frac{1}{2},\quad P(X = 2) = \frac{120}{360} = \frac{1}{3},\quad P(X = 6) = \frac{60}{360} = \frac{1}{6} \)
The expectation is
\( \displaystyle E(X) = 1 \times \frac{1}{2} + 2 \times \frac{1}{3} + 6 \times \frac{1}{6} \)
\( \displaystyle = \frac{3 + 4 + 6}{6} = \frac{13}{6} \)
Why the other options are incorrect
Option (1) 3 — Incorrect. This is the plain average of 1, 2 and 6, which treats the three sectors as equally likely.
Option (3) \(\displaystyle\frac{3}{2}\) — Incorrect. The expectation is \(\displaystyle\frac{13}{6}\), which is more than 2.
Option (4) \(\displaystyle\frac{7}{3}\) — Incorrect. \(\displaystyle\frac{7}{3} = \frac{14}{6}\), which is not \(\displaystyle\frac{13}{6}\).