Business, 11.03.2020 04:32 gbrightwell

Suppose Jack and Kate are at the town fair and are choosing which game to play. The first game has a bag with four marbles in it-1 red marble and 3 blue ones. The player draws one marble from the bag; if it is red, they win $20 and if it is blue, they win $1. The second game has a bag with 10 marbles in it-1 red, 4 blue, and 5 green. The player draws one marble from the bag; if it is red, they win $20; if it is blue, they win $5; and if it is green, they win $1. Both games cost $5 to play. Assume Jack will play the games that have a higher expected payoff than the cost of playing the game. Comparing the expected value of the payoff of each game to the price of $5 to play, we can conclude that Jack should.

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Answer from: NetherisIsTheQueen

Jack should play the first game as it has a net expected value of $0.75 greater than the cost to play

Explanation:

First game probability : four marbles = 1 ; red marble = 1/4 =0.25, blue marbles 3/4 = 0.75

second game probability : 10 marbles =1 ; red marbles =1/10 =0.1, blue marbles= 4/10=0.4 , green marble = 5/10 =0.5

revenue probability expected value

First game :

-Red marbles $20 0.25 $5

- Blue marbles $1 0.75 $0.75

Total expected value $5.75

costs to play -$5

Net value $0.75

Second game:

-Red marbles $20 0.1 $2

-blue marble $5 0.4 $2

- Green marble $1 0.5 $0.5

Total expected value $4.5

Costs to play -$5

Net value -$0.5

Answer from: Quest

unique selling propisition

Answer from: Quest

u = 30

continuing your method

$\frac{1}{10}$ = $\frac{1}{u}$ + $\frac{1}{15}$

subtract $\frac{1}{15}$ from both sides

$\frac{1}{u}$ = $\frac{1}{10}$ - $\frac{1}{15}$ thus

$\frac{1}{u}$ = $\frac{3}{30}$ - $\frac{2}{30}$

$\frac{1}{u}$ = $\frac{1}{30}$ ⇒ u = 30

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Suppose Jack and Kate are at the town fair and are choosing which game to play. The first game has a...