Select the correct answer from each drop-down menu. a dice game involves rolling two dice. a player who rolls a 3, 4, 10, 11, or 12 wins 5 points. a player who rolls a 5, 6, 7, 8, or 9 loses 5 points. if the expected value of playing this game is to lose , a player will points by rolling a 2.

Respuesta :

Answer:

The points by rolling a 2 has to be less than 65.

Step-by-step explanation:

We can find the points by rolling a 2 by using the probability distribution.

First we find the probability of each number. From the Table of Outcomes, we get:

  • possibility of rolling a 2 = 1 out of 36 → [tex]\displaystyle P(2)=\frac{1}{36}[/tex]
  • possibility of rolling a 3 = 2 out of 36 → [tex]\displaystyle P(3)=\frac{2}{36}[/tex]
  • possibility of rolling a 4 = 3 out of 36 → [tex]\displaystyle P(4)=\frac{3}{36}[/tex]
  • possibility of rolling a 5 = 4 out of 36 → [tex]\displaystyle P(5)=\frac{4}{36}[/tex]
  • possibility of rolling a 6 = 5 out of 36 → [tex]\displaystyle P(6)=\frac{5}{36}[/tex]
  • possibility of rolling a 7 = 6 out of 36 → [tex]\displaystyle P(7)=\frac{6}{36}[/tex]
  • possibility of rolling a 8 = 5 out of 36 → [tex]\displaystyle P(8)=\frac{5}{36}[/tex]
  • possibility of rolling a 9 = 4 out of 36 → [tex]\displaystyle P(9)=\frac{4}{36}[/tex]
  • possibility of rolling a 10 = 3 out of 36 → [tex]\displaystyle P(10)=\frac{3}{36}[/tex]
  • possibility of rolling a 11 = 2 out of 36 → [tex]\displaystyle P(11)=\frac{2}{36}[/tex]
  • possibility of rolling a 12 = 1 out of 36 → [tex]\displaystyle P(12)=\frac{1}{36}[/tex]

Since rolling a 3, 4, 10, 11 or 12 wins 5 points, whereas rolling a 5, 6, 7, 8 or 9 loses 5 points. Then:

  • probability of winning 5 points  [tex]\displaystyle=\frac{2}{36}+ \frac{3}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}[/tex]

                                                             [tex]\displaystyle=\frac{11}{36}[/tex]

  • probability of losing 5 points  [tex]\displaystyle=\frac{4}{36}+ \frac{5}{36}+\frac{6}{36}+\frac{5}{36}+\frac{4}{36}[/tex]

                                                         [tex]\displaystyle=\frac{24}{36}[/tex]

Refer to the Probability Distribution, we can find the expected value:

[tex]\boxed{\mu=\Sigma x_i\cdot p_i}[/tex]

where:

  • μ = expected value
  • [tex]x_i[/tex] = value (score)
  • [tex]p_i[/tex] = probability

Since expected value is to lose, therefore [tex]\boxed{\mu < 0}[/tex]

[tex]\mu < 0[/tex]

[tex]\Sigma x_i\cdot p_i < 0[/tex]

[tex]\displaystyle(5)\left(\frac{11}{36} \right)+(-5)\left(\frac{24}{36} \right)+n\left(\frac{1}{36} \right) < 0[/tex]

[tex]55-120+n < 0[/tex]

[tex]\bf n < 65[/tex]

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