Question #5685

1.In a raffle, there are 6 different types of prized: 20 $50 gift cards, 5 luxury gift baskets, 5 bottles of Don Perigon Champagne, 2 trips to Vegas, and 1 car and each prize is awarded individually. So far, there have been 9878 tickets purchased, at $5 per ticket, and you are the last person in line.
a.) How many tickets do you need to buy in order to have a 1% chance of winning any prize?
b.) if you buy 20 tickets, what is the probability that you will win a trip to Vegas or a car (or both)?

Expert's answer

a.) How many tickets do you need to buy in order to have a 1% chance of winning any prize?

There are M = 20+5+5+2+1 = 33 prize tickets at all.

P = (C(9878,n)-C(9878-33,n))/C(9878,n) = 0.01; here n is the number of bought tickets.

We can find n out of this formula by enumeration. We see that

(C(9878,3)-C(9878-33,3))/C(9878,3) ≈ 0.01,

so we need to buy 3 tickets to have a 1% chance of winning any prize.

b.) if you buy 20 tickets, what is the probability that you will win a trip to Vegas or a car (or both)?

The probability that you will win a trip to Vegas:

P1 = (C(9878,20)-C(9878-2,20))/C(9878,20), since there are two trips to Vegas.

The probability that you will win a car:

P2 = (C(9878,20)-C(9878-1,20))/C(9878,20), since there is only one car.

So, the probability that you will win a trip to Vegas or a car (or both) is

P = P1 + P2 = (2*C(9878,20)-C(9878-2,20)-C(9878-1,20))/C(9878,20) ≈ 0.6%.

There are M = 20+5+5+2+1 = 33 prize tickets at all.

P = (C(9878,n)-C(9878-33,n))/C(9878,n) = 0.01; here n is the number of bought tickets.

We can find n out of this formula by enumeration. We see that

(C(9878,3)-C(9878-33,3))/C(9878,3) ≈ 0.01,

so we need to buy 3 tickets to have a 1% chance of winning any prize.

b.) if you buy 20 tickets, what is the probability that you will win a trip to Vegas or a car (or both)?

The probability that you will win a trip to Vegas:

P1 = (C(9878,20)-C(9878-2,20))/C(9878,20), since there are two trips to Vegas.

The probability that you will win a car:

P2 = (C(9878,20)-C(9878-1,20))/C(9878,20), since there is only one car.

So, the probability that you will win a trip to Vegas or a car (or both) is

P = P1 + P2 = (2*C(9878,20)-C(9878-2,20)-C(9878-1,20))/C(9878,20) ≈ 0.6%.

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