Question

A board game uses a spinner to determine how many spaces a player will move forward on each turn. The probability is 1/2 that the player moves forward 1 space, and moving forward 2 or 3 spaces each have 1/4 probability. What is the expected value for the number of spaces a player moves forward on a turn?

Answer 1

ANSWER

1.75

EXPLANATION

The expected value of an event X is the sum of the products of each value of the event and the probability of the event resulting in that value.

In this case,

$E\lbrack X\rbrack=1\cdot P(1\text{ }forward)+2\cdot P(2\text{ }forward)+3\cdot P(3\text{ }forward)=1\cdot\frac{1}{2}+2\cdot\frac{1}{4}+3\cdot\frac{1}{4}$

Solve,

$E\lbrack X\rbrack=\frac{1}{2}+\frac{1}{2}+\frac{3}{4}=\frac{7}{4}=1.75$

Hence, the expected number of spaces the player moves in a turn is 1.75.