SOLUTION: Please please help me on this question! I have no clue how to do it. Raffle tickets numbered 1 through 30 are placed in a box. Tickets for a second raffle numbered 21 to 48 are

Algebra ->  Probability-and-statistics -> SOLUTION: Please please help me on this question! I have no clue how to do it. Raffle tickets numbered 1 through 30 are placed in a box. Tickets for a second raffle numbered 21 to 48 are       Log On


   



Question 1024565: Please please help me on this question! I have no clue how to do it.
Raffle tickets numbered 1 through 30 are placed in a box. Tickets for a second raffle numbered 21 to 48 are placed in another box. One ticket is randomly drawn from each box. What is the probability that both tickets will be greater than 20 or less than 30?
- (28/81)
- (13/10)
- (1/10)
- (361/810)

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming you mean greater than 20 and less than 30.
In the first box, there are 9 tickets (21,22,23,24,25,26,27,28,29) that meet the requirement.
So the probability is,
P%5B1%5D=9%2F%2830-1%2B1%29=9%2F30=3%2F10
In the second box, there are also 9 tickets(21,22,23,24,25,26,27,28,29) so that probability is,
P%5B2%5D=9%2F%2848-21%2B1%29=9%2F28
These are independent events so,
P=P%5B1%5D%2AP%5B2%5D
P=%281%2F30%29%289%2F28%29
P=3%2F280
Well it doesn't match any of the answers.
Please re-check the problem setup and repost.