Suppose you went to a fundraiser and you saw an interesting game. The game requires a player to select five balls from an urn that contains 1000 red balls and 4000 green balls. It costs $20 to play and the payout is as follows:
Number of red balls | Prize |
0 | $0 |
1 | $20 |
2 | $25 |
3 | $50 |
4 | $100 |
5 | $200 |
Questions:
1) Is this a binomial experiment? If so, explain what p, q, n, and x represent and find their values.
2) Someone tells you that there's 1/5 chance of getting one red ball in your 5 draws, so you can at least get your money back or win some big prizes. There's nothing to lose. Is this true? What is your expected winning in a long run? (Hint: first calculate probability of drawing 0, 1, 2, 3, 4, and 5 red balls using binomial formula and then calculate the expected value.)
1)
This is not a binomial experiment since the probability of getting a red ball changes when each of the ball is picked up.
Probability of getting red ball in first draw = 1000/5000 = 1/5
Probability of getting red ball in second draw is dependent on the first draw
2) Let X denote the number of red red b
P(X = 0) = = 0.3275
P(X = 1) = = 0.4098
P(X = 2) = 0.20485
P(X = 3) = 0.05114
P(X = 4) = 0.00637
P(X = 5) = 0.00032
Thus, Expected winning in long run = $0*0.3275 + $20*0.4098 + $25*0.20485 + $50*0.05114 + $100*0.00637 + $200*0.00032 - $20
= $-3 .425
It's not true that there's nothing to lose
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