What is the probability that the organizers win?
$begingroup$
A local fraternity is conducting a raffle where 50 tickets are to be sold - one per customer. There are three prizes to be awarded. If the four organizers of the raffle each buy one ticket, what is the probability that the four organizers win exactly two prizes?
I want to do this using combinations. I know that the probability of them winning all three of the prizes equals $dfrac{4 choose 3}{50 choose 3} = dfrac{1}{4900}$. So intuitively, I'd guess that the probability of them winning two prizes is $dfrac{4 choose 2}{binom{50}{2} times binom{3}{2}}$, the $binom{3}{2}$ being because there are 3 tickets to choose from in total. But this is apparently wrong. Why?
probability
asked Sep 14 '16 at 7:34
MaximeMaxime
153
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add a comment |
$begingroup$
A local fraternity is conducting a raffle where 50 tickets are to be sold - one per customer. There are three prizes to be awarded. If the four organizers of the raffle each buy one ticket, what is the probability that the four organizers win exactly two prizes?
I want to do this using combinations. I know that the probability of them winning all three of the prizes equals $dfrac{4 choose 3}{50 choose 3} = dfrac{1}{4900}$. So intuitively, I'd guess that the probability of them winning two prizes is $dfrac{4 choose 2}{binom{50}{2} times binom{3}{2}}$, the $binom{3}{2}$ being because there are 3 tickets to choose from in total. But this is apparently wrong. Why?
probability
asked Sep 14 '16 at 7:34
MaximeMaxime
153
$endgroup$
add a comment |
$begingroup$
A local fraternity is conducting a raffle where 50 tickets are to be sold - one per customer. There are three prizes to be awarded. If the four organizers of the raffle each buy one ticket, what is the probability that the four organizers win exactly two prizes?
I want to do this using combinations. I know that the probability of them winning all three of the prizes equals $dfrac{4 choose 3}{50 choose 3} = dfrac{1}{4900}$. So intuitively, I'd guess that the probability of them winning two prizes is $dfrac{4 choose 2}{binom{50}{2} times binom{3}{2}}$, the $binom{3}{2}$ being because there are 3 tickets to choose from in total. But this is apparently wrong. Why?
probability
asked Sep 14 '16 at 7:34
MaximeMaxime
153
$endgroup$
A local fraternity is conducting a raffle where 50 tickets are to be sold - one per customer. There are three prizes to be awarded. If the four organizers of the raffle each buy one ticket, what is the probability that the four organizers win exactly two prizes?
I want to do this using combinations. I know that the probability of them winning all three of the prizes equals $dfrac{4 choose 3}{50 choose 3} = dfrac{1}{4900}$. So intuitively, I'd guess that the probability of them winning two prizes is $dfrac{4 choose 2}{binom{50}{2} times binom{3}{2}}$, the $binom{3}{2}$ being because there are 3 tickets to choose from in total. But this is apparently wrong. Why?
probability
probability
asked Sep 14 '16 at 7:34
MaximeMaxime
153
asked Sep 14 '16 at 7:34
MaximeMaxime
153
asked Sep 14 '16 at 7:34
MaximeMaxime
153
asked Sep 14 '16 at 7:34
MaximeMaxime
153
asked Sep 14 '16 at 7:34
MaximeMaxime
153
153
add a comment |
add a comment |
1 Answer
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$begingroup$
But this is apparently wrong. Why?
You have calculated the probability of selecting two from four winning tickets, out of all ways to select two from fifty tickets and two from other three things. This is not ... like anything that you want to find.
You want the probability of selecting two from four winning tickets and one from forty-six losing tickets, out of all the ways to select any three from all fifty tickets.
answered Sep 14 '16 at 7:45
Graham KempGraham Kemp
85.8k43378
$endgroup$
$begingroup$
Ah, I overlooked that technique, it sounds much more logical. But could you explain what is exactly wrong about my idea?
$endgroup$
– Maxime
Sep 14 '16 at 7:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
But this is apparently wrong. Why?
You have calculated the probability of selecting two from four winning tickets, out of all ways to select two from fifty tickets and two from other three things. This is not ... like anything that you want to find.
You want the probability of selecting two from four winning tickets and one from forty-six losing tickets, out of all the ways to select any three from all fifty tickets.
answered Sep 14 '16 at 7:45
Graham KempGraham Kemp
85.8k43378
$endgroup$
$begingroup$
Ah, I overlooked that technique, it sounds much more logical. But could you explain what is exactly wrong about my idea?
$endgroup$
– Maxime
Sep 14 '16 at 7:46
add a comment |
$begingroup$
But this is apparently wrong. Why?
You have calculated the probability of selecting two from four winning tickets, out of all ways to select two from fifty tickets and two from other three things. This is not ... like anything that you want to find.
You want the probability of selecting two from four winning tickets and one from forty-six losing tickets, out of all the ways to select any three from all fifty tickets.
answered Sep 14 '16 at 7:45
Graham KempGraham Kemp
85.8k43378
$endgroup$
$begingroup$
Ah, I overlooked that technique, it sounds much more logical. But could you explain what is exactly wrong about my idea?
$endgroup$
– Maxime
Sep 14 '16 at 7:46
add a comment |
$begingroup$
But this is apparently wrong. Why?
You have calculated the probability of selecting two from four winning tickets, out of all ways to select two from fifty tickets and two from other three things. This is not ... like anything that you want to find.
You want the probability of selecting two from four winning tickets and one from forty-six losing tickets, out of all the ways to select any three from all fifty tickets.
answered Sep 14 '16 at 7:45
Graham KempGraham Kemp
85.8k43378
$endgroup$
But this is apparently wrong. Why?
You have calculated the probability of selecting two from four winning tickets, out of all ways to select two from fifty tickets and two from other three things. This is not ... like anything that you want to find.
You want the probability of selecting two from four winning tickets and one from forty-six losing tickets, out of all the ways to select any three from all fifty tickets.
answered Sep 14 '16 at 7:45
Graham KempGraham Kemp
85.8k43378
edited Sep 14 '16 at 7:52
answered Sep 14 '16 at 7:45
Graham KempGraham Kemp
85.8k43378
answered Sep 14 '16 at 7:45
Graham KempGraham Kemp
85.8k43378
answered Sep 14 '16 at 7:45
Graham KempGraham Kemp
85.8k43378
85.8k43378
$begingroup$
Ah, I overlooked that technique, it sounds much more logical. But could you explain what is exactly wrong about my idea?
$endgroup$
– Maxime
Sep 14 '16 at 7:46
add a comment |
$begingroup$
Ah, I overlooked that technique, it sounds much more logical. But could you explain what is exactly wrong about my idea?
$endgroup$
– Maxime
Sep 14 '16 at 7:46
$begingroup$
Ah, I overlooked that technique, it sounds much more logical. But could you explain what is exactly wrong about my idea?
$endgroup$
– Maxime
Sep 14 '16 at 7:46
$begingroup$
Ah, I overlooked that technique, it sounds much more logical. But could you explain what is exactly wrong about my idea?
$endgroup$
– Maxime
Sep 14 '16 at 7:46
add a comment |
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