Bradford T. answered • 05/13/21
Tutor
4.9
(29)
MS in Electrical Engineering with 40+ years as an Engineer
The payout to winners is:
1,000,000 + 25,000 + 5(1000) = $1,030,000
If 2,000,000 tickets are sold at $2 each. The income before paying the winners = $4,000,000
So the overall profit is 4,000,000 - 1,030,000 = $2970000
The profit per ticket = $2970000/2000000 = $1.485
If I understand the problem correctly.
Mark Z.
Suppose that W₁ represents the first winning ticket and P(W₁) is the probability of holding that ticket. For the 1st prize, there is 1 winner out of 2×10⁶ ticket holders so P(1st) = 1/(2×10⁶) = 5× 10⁻⁷, where P(1st) = P(W₁) is the probability of winning 1st prize (equivalent to holding the W₁ticket) . When it comes to 2nd prize, there is still one winner. However, only (2×10⁶ − 1) ticket-holders remain, since one choice was “used-up” when the 1st place winer was chosen. So, we get P(2nd) = P(W₂) = 1/(2×10⁶ − 1) or P(2nd) ≈ 5× 10⁻⁷ since 2×10⁶ and (2 × 10⁶ − 1) are approximately equal. The next five winners are added together because all five count as 3rd place so P(3rd) = P(W₃) + P(W₄) + P(W₅) + P(W₆) + P(W₇). For each person in 3rd place, one ticket is drawn but the number of remaining tickets is reduced by one for each selection: P(W₃) = 1/(2×10⁶ − 2) and P(W₄) = 1/(2×10⁶ − 3) and P(W₅) = 1/(2×10⁶ − 4) and P(W₆) = 1/(2×10⁶ − 5) and P(W₇) = 1/(2×10⁶ − 6). The difference between each denominator and 2×10⁶ is negligible so P(3rd) ≈ 5/(2×10⁶) ≈ 2.5×10⁻⁶ since P(3rd) is a sum of 5 (nearly) identical fractions. For a valid probability distribution, the chance of loosing must be whatever remains when people who won are subtracted: P(L) = 1 - P(1st) - P(2nd) - P(3rd) = 0.9999985, where P(L) is the probability of losing. Suppose that K is the amount of money "kept" by the lottery organizers with K₁ = (2 -1×10⁶) for 1st prize, K₂ = (2-25× 10³) for 2nd prize, K₃ = (2-1× 10³) for 3rd prize and K₄₊ = 2 for all the losers. To answer the question, we must calculate the expectation value: E[K] = K₁×P(1st) + K₂×P(2nd) + K₃×P(3rd) + K₄₊×P(L) ≈ 1.485004 (using the numbers from earlier steps)
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08/14/24
Lia K.
Thank you for the help, So my answer was correct? For this problem your supposed to use the expected value formula, so you have to find the probability for each ticket then add all of the possibilities together.05/13/21