In the JackRat lottery, three white balls are drawn (at random) from ten balls numbered from 11 to 20, and one blue AwesomeBall is drawn (at random) from ten balls numbered from 21 to 30. When you buy a ticket, you select three numbers from 11 to 20 and one number from 21 to 30. To win an awesome prize, the numbers on your ticket must match at least two of the white balls or must match the blue AwesomeBall.

If you buy a ticket, what is your probability of winning an awesome prize?

Forumofweb2.0cal Mar 16, 2020

#1**0 **

On average, 1 out of 10 times you will draw the AwesomeBall (and this wins all by itself so we don't care what the other balls are) for a winning probability of 0.10.

Now, for the 9 out of ten times that you do not draw the AwesomeBall:

We will have to get a losing blue ball, I'll call that B (and the probability of getting that losing ball is 0.90).

We can win by getting two or more of the white balls. I'll classify a winning white ball (whose probability is 1/10 = 010)

as W and a losing white ball as L (whose probability is 9/10 = 0.90).

So, we can still win if we have this:

B x W x W x W = (0.90) x (0.10) x (0.10) x (0.10) = 0.0009

B x W x W x L = (0.90) x (0.10) x (0.90) x (0.10) = 0.0081

B x W x L x W = ...

B x L x W x W = ...

Since these are independent, add the above 5 results together: 0.10 + 0.009 + 0.0081 + ... + ... = answer

geno3141 Mar 16, 2020

#4**0 **

Yeah, I messed up my answer.

Getting the AwesomeBall is still 0.10 and, if we get that ball, we have won.

Now, what happens when we don't get that ball (which is a probability of 0.90).

B x W x W x W = (0.90) x (3/10) x (2/9) x (1/8) = 0.0075

B x W x W x L = (0.90) x (3/10) x (2/9) x (7/8) =

B x W x L x W = (0.90) x (3/10) x (7/9) x (2/8) =

B x L x W x W = (0.90) x (7/10) x (3/9) x (2/8) =

Now, add 0.10 + 0.0075 + ...

geno3141 Mar 19, 2020