Answer to Question #5014 in Statistics and Probability for Ainy

Question #5014
If I buy 4 tickets in a 500 ticket lottery, determine the probability that I win: a) the first 3 prizes b) at least one of the 3 prizes.
Let&#039;s choose a ticket that will not take value more than 3. That can be made in
C(4,1)=4ways. Also, it can take values from 4 to 500, in fact 500-4=496 different values.
Then, let&#039;s choose the values for the other tickets from the set of {1,2,3}. That can be made in
C(3,1)*C(2,1)*C(1,1) = 3! = 6.
ways. You see that it is the number of permutations of these tickets.
So, the total number of possible favorable cases is
N = 496*4*6 = 11904.

The total number of different possible cases is
M = C(500,1)*C(499,1)*C(498,1)*C(497,1) = 500*499*498*497 = 61752747000.

So, the probability of getting the first 3 prizes is
P = N/M = 11904/61752747000 &asymp; 1,92768*10^(-7)

b) at least one of the 3 prizes
Let&#039;s calculate the number of favorable cases:
N = C(4,1)*C(3,1)*499*498*497 = 4*3*499*498*497 = 1482065928
Here we choose one ticket from four ones that will take the prize number, and then choose its value from {1,2,3}.
The total number of possible favorable cases is the same. So,
P = N/M = 1482065928/61752747000 = 0,024 = 2.4%.

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