Probability of objects are being sorted
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There are 20 toffees (10 of mint, 10 of orange) in two boxes. I mix them together. Then I put 10 randomly chosen from the pot into one box and 10 in the other box. What is the probability that I end up with 10 orange toffees in one box and 10 mint toffees in the other?
probability probability-theory elementary-probability
asked Dec 3 at 17:58
Aashish
1847
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up vote
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There are 20 toffees (10 of mint, 10 of orange) in two boxes. I mix them together. Then I put 10 randomly chosen from the pot into one box and 10 in the other box. What is the probability that I end up with 10 orange toffees in one box and 10 mint toffees in the other?
probability probability-theory elementary-probability
asked Dec 3 at 17:58
Aashish
1847
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
There are 20 toffees (10 of mint, 10 of orange) in two boxes. I mix them together. Then I put 10 randomly chosen from the pot into one box and 10 in the other box. What is the probability that I end up with 10 orange toffees in one box and 10 mint toffees in the other?
probability probability-theory elementary-probability
asked Dec 3 at 17:58
Aashish
1847
There are 20 toffees (10 of mint, 10 of orange) in two boxes. I mix them together. Then I put 10 randomly chosen from the pot into one box and 10 in the other box. What is the probability that I end up with 10 orange toffees in one box and 10 mint toffees in the other?
probability probability-theory elementary-probability
probability probability-theory elementary-probability
asked Dec 3 at 17:58
Aashish
1847
asked Dec 3 at 17:58
Aashish
1847
asked Dec 3 at 17:58
Aashish
1847
asked Dec 3 at 17:58
Aashish
1847
asked Dec 3 at 17:58
Aashish
1847
1847
add a comment |
add a comment |
2 Answers
2
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1
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There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 at 18:04
Ofya
4798
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up vote
1
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You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 at 18:06
Sean Lee
1387
Shouldn't answer be simply 1/2
– Aashish
Dec 4 at 1:09
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
– Sean Lee
Dec 4 at 2:36
Yes got my mistake...thanx !!
– Aashish
Dec 4 at 9:03
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 at 18:04
Ofya
4798
add a comment |
up vote
1
down vote
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 at 18:04
Ofya
4798
add a comment |
up vote
1
down vote
up vote
1
down vote
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 at 18:04
Ofya
4798
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 at 18:04
Ofya
4798
answered Dec 3 at 18:04
Ofya
4798
answered Dec 3 at 18:04
Ofya
4798
answered Dec 3 at 18:04
Ofya
4798
4798
add a comment |
add a comment |
up vote
1
down vote
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 at 18:06
Sean Lee
1387
Shouldn't answer be simply 1/2
– Aashish
Dec 4 at 1:09
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
– Sean Lee
Dec 4 at 2:36
Yes got my mistake...thanx !!
– Aashish
Dec 4 at 9:03
add a comment |
up vote
1
down vote
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 at 18:06
Sean Lee
1387
Shouldn't answer be simply 1/2
– Aashish
Dec 4 at 1:09
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
– Sean Lee
Dec 4 at 2:36
Yes got my mistake...thanx !!
– Aashish
Dec 4 at 9:03
add a comment |
up vote
1
down vote
up vote
1
down vote
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 at 18:06
Sean Lee
1387
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 at 18:06
Sean Lee
1387
answered Dec 3 at 18:06
Sean Lee
1387
answered Dec 3 at 18:06
Sean Lee
1387
answered Dec 3 at 18:06
Sean Lee
1387
1387
Shouldn't answer be simply 1/2
– Aashish
Dec 4 at 1:09
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
– Sean Lee
Dec 4 at 2:36
Yes got my mistake...thanx !!
– Aashish
Dec 4 at 9:03
add a comment |
Shouldn't answer be simply 1/2
– Aashish
Dec 4 at 1:09
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
– Sean Lee
Dec 4 at 2:36
Yes got my mistake...thanx !!
– Aashish
Dec 4 at 9:03
Shouldn't answer be simply 1/2
– Aashish
Dec 4 at 1:09
Shouldn't answer be simply 1/2
– Aashish
Dec 4 at 1:09
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
– Sean Lee
Dec 4 at 2:36
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
– Sean Lee
Dec 4 at 2:36
Yes got my mistake...thanx !!
– Aashish
Dec 4 at 9:03
Yes got my mistake...thanx !!
– Aashish
Dec 4 at 9:03
add a comment |
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