100 PNTS AND BRAINLIST FOR CORRECT ANWSER WITH A GOOD EXPLAINATION!! You pay $10 to play the following game of chance. There is a bag containing 20 balls: five are red, five are green, and the rest are yellow. You are to draw one ball from the bag. You will win $18 if you draw a red ball, and you will win $10 if you draw a green ball. You win nothing for drawing a yellow ball. What is the expected value? Should you play?

your answer is 20

Step-by-step explanation:

Expert's answer

Because the events (draw a red/green/yellow ball) form the full group and do not intersect, we can calculate the expected win/lose for 1 game as

\mathbb{E}X = \sum\limits_k {{w_k}P(k)}EX=k∑wkP(k)

Where {P(k)}P(k) is a probability of the certain event, that can be calculated as P(k) = \frac{{{N_k}}}{N}P(k)=NNk where {N_k}Nk is the number of k-type balls and N = \sum\limits_k {{N_k}}N=k∑Nk is the amount of balls, {{w_k}}wk is the win/lose if we get the k-th type ball (it can be calculated as {w_k} = {x_k} - ywk=xk−y where {x_k}xk is the win for the k-th type ball and yy is the price for the 1 game). We get

\mathbb{E}X = \sum\limits_k {{w_k}P(k)} = \$4 \cdot \frac{5}{{12}} +\$ 2 \cdot \frac{4}{{12}} + \$( - 10) \cdot \frac{3}{{12}} = -\$ \frac{1}{6}EX=k∑wkP(k)=$4⋅125+$2⋅124+$(−10)⋅123=−$61

Then because the games are independent, the total result after 100 times is

\mathbb{E}Y = n \cdot \mathbb{E}X = - \$\frac{1}{6} \cdot 100 = - \$\frac{{50}}{3} \approx - \$16.667EY=n⋅EX=−$61⋅100=−$350≈−$16.667

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step-by-step explanation: