Maths Made Easy Forums Standard Probability & Relative Frequency Financial Expectation

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Keymaster
Post count: 659

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 • Admin Keymaster Post count: 659 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 • Admin Keymaster Post count: 659 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 • Admin Keymaster Post count: 659 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

There is a game where there is a 20% chance of winning $50, 50% chance of winning$5 or losing $5 if no prize is won. What is the financial expectation of this game? P(lose) = 0.3 E = 0.20 ×$50 + 0.50 × $5 + 0.30 × -$5