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Tom claims that there is over a 60% chance that you can win a 10 or 5 prize. Is his claim correct


New member
Apr 13, 2020
hello! I have two questions I needed some help with.

2. Tom is having a 4/19 lottery as a fundraiser; 4 winning numbers will be randomly selected without replacement out of the numbers from 1 to 19 inclusive. (similar to Ontario’s 6/49, with smaller numbers) The prizes are described below. For the questions below, define an appropriate random variable X.
Tickets Prize
All 4 winning numbers 1000 dollars
Any 3 of the winning numbers 100
Any 2 of the winning numbers 10
Any 1 of the winning numbers 5
None of the winning numbers 0
a) Tom claims that there is over a 60% chance that you can win a $10 or $5 prize. Is his claim correct?
b) What would be an appropriate cost per ticket. Explain your answer.
c) Given the cost suggested in part b), what is his expected profit on 100 tickets?
Note: Order doesn't matter for the numbers

I solved part a and got
p(X=4)= 1/19C4= 1/3876
p(X=3)= (4C3 * 15C1)/3876= 60/3876
p(X=2)= (4C2 * 15C2)/3876= 630/3876
p(X=1)= (4C1 * 15C3)/3876= 1820/3876
a. (630 + 1820)/3876≈ 0.632. which is 63.2%, so yes his claim is correct.

I needed help with b and c. For b, I know I should calculate expectations and then choose a price that gives a good revenue based on payout but I wasn't sure how.

3. A discrete random variable, X, is uniformly distributed between 1 and 10 inclusive and another discrete random variable, Y, is uniformly distributed between 5 and 15 inclusive. If one number from each set is selected at random, then determine P(Y < X ) .

I solved it the following:
(15 cases in which Y < X) ÷ (110 possible pairs) = 3/22

I got 15 cases because if Y is 5, X can be 6-10 (5 options), then if Y is 6, X can be 7-10 (4 options), if Y is 7, X can be 8-10 (3 options), if Y is 8, X can be 9 or 10 (2 options) and if Y is 9, X can be 10 ( 1 option).
5 +4+3+2+1= 15 cases in which Y < X)
and 11 X values x 10 Y values = 110 possible pairs

But I feel like it's the wrong approach as it doesn't use any probability distribution formula so I was looking for some help on it too.
Thanks in advance!


Well-known member
MHB Math Helper
Jan 29, 2012
For (b), to determine how much should be charged for a ticket, you idea would be correct if this were to be a "break even" event- all money brought in going out as prizes. But it isn't, it is a "fund-raiser"! Without knowing how much money is expected to be raised, it is impossible to say how much should be charged per ticket.