Question

I am working on a question for Probability that is dealing with discrete random variables and their probability distributions and I could use some assistance in figuring out how to solve it. I have dealt with similar problems already, but for some reason this one keeps tripping me up.

Answer 1

The question itself is:
"Four possibly winning numbers for a lottery—AB-4536, NH-7812, SQ-7855, and ZY-3221— arrive in the mail. You will win a prize if one of your numbers matches one of the winning numbers contained on a list held by those conducting the lottery. One first prize of $100,000, two second prizes of $50,000 each, and ten third prizes of $1000 each will be awarded. To be eligible to win, you need to mail the coupon back to the company at a cost of 33¢ for postage. No purchase is required. From the structure of the numbers that you received, it is obvious the numbers sent out consist of two letters followed by four digits. Assuming that the numbers you received were generated at random, what are your expected winnings from the lottery? Is it worth 33¢ to enter this lottery?"

Thus far I have figured out that the total sample points for the lottery numbers is 68,250, but I am unsure how to proceed from there.

Answer 2

I think there are:
\[{26^2} \cdot {10^4} = 6,760,000\]
possible combinations

Answer 3

Uff da, I was way off then with that first part then. I had been using 26C2 and 10C4. Thank you. How would you recommend figuring out the probability for winning a prize? I am not sure if to approach it. I was figuring the numerator would be the product of 13C4 and some others, but I was not sure if it would be 1C1, 2C2, and 10C4 or something else.

Answer 4

And would you recommend using a binomial or a hypergeometric distribution for it?

Answer 5

I think that the probability \(p\), of win, for a single player, is:
\(p=\)favorable outcomes over possible outcomes, or:
\[p = \frac{4}{{{{26}^2} \cdot {{10}^4}}}\]

Answer 6

I mean the probability to win at least one prize

Answer 7

Oh boy, I was definitely overthinking the whole thing then and trying to include the small details. Thank you very much for the assistance with the problem, I know how to take the rest from that point then.

Answer 8

:) please, I think that we can apply the binomial distribution

Answer 9

Alright, will do, thank you!

Answer 10

:)