Respuesta :
Answer:
The expected value of the points earned on a single roll in this game is [tex]\dfrac{1}{6} = 0.1667[/tex] .
Step-by-step explanation:
We are given that consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number, that many points will be added to the player’s total. Any other roll will be deducted from the player’s total.
Assuming that the numbered cube is a dice with numbers (1, 2, 3, 4, 5, and 6).
Here, the prime numbers are = 1, 2, 3 and 5
Numbers which are not prime = 4 and 6
This means that if the dice got the number 1, 2, 3 or 5, then that many points will be added to the player’s total and if the dice got the number 4 or 6, then that many points will get deducted from the player’s total.
Here, we have to make a probability distribution to find the expected value of the points earned on a single roll in this game.
Note that the probability of getting any of the specific number on the dice is [tex]\dfrac{1}{6}[/tex] .
Numbers on the dice (X) P(X)
+1 [tex]\frac{1}{6}[/tex]
+2 [tex]\frac{1}{6}[/tex]
+3 [tex]\frac{1}{6}[/tex]
-4 [tex]\frac{1}{6}[/tex]
+5 [tex]\frac{1}{6}[/tex]
-6 [tex]\frac{1}{6}[/tex]
Here (+) sign represent the addition in the player's total and (-) sign represents the deduction in the player's total.
Now, the expected value of X, E(X) = [tex]\sum X \times P(X)[/tex]
= [tex](+1) \times \frac{1}{6} +(+2) \times \frac{1}{6} +(+3) \times \frac{1}{6} +(-4) \times \frac{1}{6} +(+5) \times \frac{1}{6} +(-6) \times \frac{1}{6}[/tex]
= [tex]\frac{1}{6} + \frac{2}{6} + \frac{3}{6} - \frac{4}{6} + \frac{5}{6} - \frac{6}{6}[/tex]
= [tex]\frac{1+2+3-4+5-6}{6}[/tex]
= [tex]\frac{11-10}{6}= \frac{1}{6}[/tex]
Hence, the expected value of the points earned on a single roll in this game is [tex]\frac{1}{6} = 0.1667[/tex] .
