Question

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

Answer 1

its 1/4

.......................

re:zero

Answer 2

Answer:

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

           A.    3 spaces forward.

Step-by-step explanation:

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:

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

                 Hence, the answer is:

              A.  3 spaces forward.