In a particular game, a spinner with four equally-sized sectors labeled 1, 4, 6, and 8 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. 3 spaces forward
B. 3 spaces backward
C. 1 space backward
D. 1 space forward

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virtuematane virtuematane · Answer 1 · 2019-06-07T07:59:48+02:00

The mathematical expectation for the number of spaces moved in one turn is:

A. 3 spaces forward.

Th result or the sample space on spinning a spinner twice is:

(1,1) (1,4) (1,6) (1,8)

(4,1) (4,4) (4,6) (4,8)

(6,1) (6,4) (6,6) (6,8)

(8,1) (8,4) (8,6) (8,8)

Total number of outcomes= 16

The number of outcomes whose sum is even= 10

( Since the outcomes are: {(1,1) , (4,4) , (4,6) , (4,8) , (6,4) , (6,6) , (6,8) , (8,4) , (8,6) , (8,8)} )

The number of outcomes whose sum is odd= 6

( Since, the outcomes are: { (1,4) , (1,6) , (1,8) , (4,1) , (6,1) , (8,1) }

Probability(sum even)=10/16

Probability(sum odd)=6/16

Hence, the expectation is:

[tex]E(X)=\dfrac{10}{16}\times (+6)+\dfrac{6}{16}\times (-2)\\\\\\E(X)=\dfrac{60-12}{16}\\\\\\E(X)=\dfrac{48}{16}\\\\\\E(X)=+3[/tex]

Hence, the answer is:

A. 3 spaces forward.