Answered

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A 6-sided die is loaded in such a way where odd outcomes occur 2/9 of the time, and even outcomes happen the other
1/9 of the time. What is this die's expected value?
A. 3.25
B. 4
C. 3.5
D. 3.33

Sagot :

Answer:

3.33, or 3 1/3

Step-by-step explanation:

Expected value:

[tex](9 \times \frac{2}{9} ) + (12 \times \frac{1}{9} ) = 2 + \frac{4}{3} = 2 + 1 \frac{1}{3} = 3 \frac{1}{3} [/tex]

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