+0  
 
0
216
2
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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?

 

3 spaces forward

1 space backward

3 spaces backward

1 space forward

 May 5, 2020
 #1
avatar+50 
+1

i believe the answer is 0 for all of them, since the only amount of spaces that you can move is even, and an even number + even number = even number and even number - even number = even number

 

edit: aaaaaah. misunderstood the problem. CPhill got it right!

 May 5, 2020
edited by chrissy  May 5, 2020
 #2
avatar+111326 
+2

        1     4     6      8

1      2     5     7      9

4      5     8     10   12

6      7     10   12   14

8      9     12   14    16

 

P(even sum) =  10/16  =  5/8

P(odd sum) = 6/16    =  3/8

 

Expected value

 

(5/8) (6)  +   (3/8) (-2)  =

 

30/8  - 6/8  =

 

24/8  =

 

3 spaces forward

 

 

cool cool cool

 May 5, 2020

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