In a particular game, a spinnerwith four equally-sized sectors labeled 1, 3, 4, and 6 is spun twice. one turn is considered 2 spins of the spinner. if the sum of the spins is even, you move forward 6 spaces. otherwise, you move back 2 spaces. what is the mathematical expectation for the number of spaces moved in one turn?

a. 2 spaces backward

b. 2 spaces forward

c. 1 space backward

d. 1 space forward

Each spin consists in a choice of a number among 1, 3, 4 and 6. This means that there are two odd numbers and two even numbers.

You get an even number if you sum two numbers with the same parity, you get an odd number if you sum two numbers with opposite parity.

Whatever number you get from the first spin, you have a 50% chance of spinning a number with the same parity with the second spin, and a 50% chance of spinning a number with opposite parity with the second spin.

So, with each turn, you have a 50% chance of getting an even sum, and a 50% chance of getting an odd sum.

This means that, by definition, the expected number of spaces moved is

So, on average, you move 2 steps foward.

answer: hey there,

the answer you are looking for is a..

hope this you!

so, converting firstly the mixed fractions to improper fractions and then "adding".