MATH SOLVE

2 months ago

Q:
# What is the expected value of the game if 3 hits pay $6, 2 hits pays $4, 1 hit pays $2, and 0 hits pays costs $4?

Accepted Solution

A:

The probability of getting 3 hits is

[tex]P\left(3\:hits\right)=0.7\times 0.5\times 0.3=0.105[/tex]

The probability of getting 2 hits is

[tex]P\left(2\:hits\right)=\left(0.7\times 0.5\times 0.7\right)+\left(0.7\times 0.5\times 0.3\right)+\left(0.3\times 0.5\times 0.3\right)=0.395[/tex]

The probability of getting 1 hit is

[tex]P\left(1\:hit\right)=\left(0.7\times 0.5\times 0.7\right)+\left(0.3\times 0.5\times 0.7\right)+\left(0.3\times 0.5\times 0.3\right)=0.395[/tex]

The probability of getting 0 hit is

[tex]P\left(0\:hit\right)=0.3\times 0.5\times 0.7=0.105[/tex]

The expected value is solved by adding the products of the probability by the given pay. That is

[tex]E=0.105\left(6\right)+0.395\left(4\right)+0.395\left(2\right)+0.105\left(4\right)=3.42[/tex]

The expected value is $3.42.

[tex]P\left(3\:hits\right)=0.7\times 0.5\times 0.3=0.105[/tex]

The probability of getting 2 hits is

[tex]P\left(2\:hits\right)=\left(0.7\times 0.5\times 0.7\right)+\left(0.7\times 0.5\times 0.3\right)+\left(0.3\times 0.5\times 0.3\right)=0.395[/tex]

The probability of getting 1 hit is

[tex]P\left(1\:hit\right)=\left(0.7\times 0.5\times 0.7\right)+\left(0.3\times 0.5\times 0.7\right)+\left(0.3\times 0.5\times 0.3\right)=0.395[/tex]

The probability of getting 0 hit is

[tex]P\left(0\:hit\right)=0.3\times 0.5\times 0.7=0.105[/tex]

The expected value is solved by adding the products of the probability by the given pay. That is

[tex]E=0.105\left(6\right)+0.395\left(4\right)+0.395\left(2\right)+0.105\left(4\right)=3.42[/tex]

The expected value is $3.42.