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Amy buys a $1 ticket in a raffle. There are 200 tickets in the raffle and two prizes. First prize is $100 and second prize is $50. Find Amy’s financial expectation.
1st prize = ^{1}/_{200 }× 100
= $0.50
2nd prize = ^{1}/_{200} × 50
= $0.25
No prize = ^{198}/_{100} × -1
= -$0.99
total = 50 + 25 – 99 = -24
∴ loss of $0.24
In a raffle, 1000 tickets are sold costing $2 each. First prize is $350, second prize is $175, third prize is $100 and fourth prize is $50. No ticket can win more than one prize. What is the probability of winning first prize? What is the expected financial return?
P(win first prize) = ^{1}/_{1000
}P(1st) = ^{1}/_{1000}
P(2nd) = ^{1}/_{1000}
P(3rd) = ^{1}/_{1000}
P(no prize) = ^{997}/_{1000}
E = ^{1}/_{1000} × 250 + ^{1}/_{1000} × 125 + ^{1}/_{1000} × 100 + ^{997}/_{1000} × -2
E = -1.519
∴ lose $1.52
A game is played by tossing two coins. The rules are: If two heads are thrown you win $10. If two heads are not thrown you lose $5.
a. Determine the financial expectation for this game.
b. If you played this game 50 times, how much would you expect to win or lose?
a. P(2 heads) = ¼ P(not 2 heads) = ¾
E = ¼ × 10 + ¾ × -5
E = -1.25
Lose $1.25
b. n = 50
E = 50 × -1.25
= -62.5
∴ lose $62.50
It costs $12 to play a game. There is a probability of $$\frac14$$ of winning $40 and a probability of $$\frac38$$ of winning $25. Calculate the financial expectation of this game.
P(lose) = $$1-\frac14-\frac38=\frac38$$
E = $$\frac14\times\,$40+\frac38\times\,$25-\frac38\times\,$10-$12$$ (don’t forget to subtract the cost of playing the game)
= $3.625
∴ E = $3.63