I would like to participate in a raffle where only 4500 tickets will be sold. There are 7 prizes available. If I purchase 6 tickets what is the probability that I would win a prize?
2007-06-16 05:56:30 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The probability of winning the first of the seven prizes is
1/4500. The probability of not winning it is 4499/4500.
If you dont win the first, the probability of winning the second is 1/4499 and the probability of not winning is 4498/4499.
If you add up all the probabilities for the seven prizes you get
1/ 4500 +
1/ 4499 +
1/ 4498 +
1/ 4497 +
1/ 4496 +
1/ 4495 +
1/ 4494 = 7/4497
or about 0.1557% chance to win one prize with one ticket.
If you have 6 tickets, your odds go up to 42/4487 or 14/1499 or about 0.9340%. Its just less than 1% or one in a hundred chance to win one prize.
Remember that they have to sell all the tickets, the selection process must be truly random, and winning tickets are not put back into the pot for the next draw.
2007-06-16 06:07:33 · answer #1 · answered by Curly 6 · 0⤊ 0⤋
Assuming 7 drawings and that you would not win -
1)4500 to 6 = 750 to 1
2)4499 to 6 = 749.83 to 1
3)4498 to 6 = 749.67 to 1
4)4497 to 6 = 749.5 to 1
5)4496 to 6 = 749.33 to 1
6)4495 to 6 = 749.17 to 1
7)4494 to 6 = 749 to 1
Overall, your odds of winning one prize would be 1 in 749.5.
2007-06-16 06:16:05 · answer #2 · answered by OkieDanCer 3 · 0⤊ 0⤋
The probability that you don't win a prize is the probability that none of your 6 tickets is a winner. This probability is
((4500-7) choose 6)/(4500 choose 6)
=(4493 choose 6)/(4500 choose 6)
=4493!/6!/4487!
/(4500!/6!/4494!)
=4493*4492*4491*4490*4489*4488
/(4495*4496*4497
*4498*4499*4500)
=0.9906977386
So the odds of winning a prize is 1-0.9906977386
=0.0093022614
2007-06-16 06:13:29 · answer #3 · answered by Anonymous · 1⤊ 0⤋
A solution to get an approximate reply: you already know the likelihood of a million participant wasting is two/three. So the likelihood of two gamers wasting is roughly (two/three)^two = four/nine. So the threat of a minimum of certainly one of them profitable is roughly five/nine that is fifty five.6%. (The motive this is not the specified reply is on the grounds that the parties are not unbiased.) To get the specified reply: P(win 1st prize) + P(leave out 1st, win second) + P(leave out 1st and second, win third) + ... = (two/30) + (28/30)(two/29) + two(28p2 / 30p3) + two(28p3 / 30p4) + ... + two(28p9 / 30p10) = forty nine/87 = fifty six.three%
2016-09-05 18:21:44 · answer #4 · answered by cinnante 4 · 0⤊ 0⤋
42 out of 4500
2007-06-16 06:02:29 · answer #5 · answered by SagarSaroj 2 · 0⤊ 0⤋
(6/4500)*7= 0.0093 = 0.93%
2007-06-16 06:07:25 · answer #6 · answered by ironduke8159 7 · 0⤊ 0⤋