Pulling colored balls from a bag
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Balls are randomly removed from a bag without replacement. If the probability that the first five balls withdrawn are all green is one-half, what is the fewest possible number of balls in the bag at the start?
I believe the answer is 10 balls because we can assume that there are 5 green balls to begin with. Does this logic hold?
probability combinatorics
asked Nov 15 at 16:29
Arthur Green
555
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up vote
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Balls are randomly removed from a bag without replacement. If the probability that the first five balls withdrawn are all green is one-half, what is the fewest possible number of balls in the bag at the start?
I believe the answer is 10 balls because we can assume that there are 5 green balls to begin with. Does this logic hold?
probability combinatorics
asked Nov 15 at 16:29
Arthur Green
555
2
No. If there are $10$ balls, $5$ of which are green, the probability that the first five are green is $1big /binom {10}5neq frac 12$.
– lulu
Nov 15 at 16:30
@lulu thank you that is what I was thinking. I will keep working on this problem
– Arthur Green
Nov 15 at 16:32
4
But if there are $10$ balls, $9$ of which are green, then the probability that the first five are green is $frac12$.
– SmileyCraft
Nov 15 at 16:32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Balls are randomly removed from a bag without replacement. If the probability that the first five balls withdrawn are all green is one-half, what is the fewest possible number of balls in the bag at the start?
I believe the answer is 10 balls because we can assume that there are 5 green balls to begin with. Does this logic hold?
probability combinatorics
asked Nov 15 at 16:29
Arthur Green
555
Balls are randomly removed from a bag without replacement. If the probability that the first five balls withdrawn are all green is one-half, what is the fewest possible number of balls in the bag at the start?
I believe the answer is 10 balls because we can assume that there are 5 green balls to begin with. Does this logic hold?
probability combinatorics
probability combinatorics
asked Nov 15 at 16:29
Arthur Green
555
asked Nov 15 at 16:29
Arthur Green
555
asked Nov 15 at 16:29
Arthur Green
555
asked Nov 15 at 16:29
Arthur Green
555
asked Nov 15 at 16:29
Arthur Green
555
555
2
No. If there are $10$ balls, $5$ of which are green, the probability that the first five are green is $1big /binom {10}5neq frac 12$.
– lulu
Nov 15 at 16:30
@lulu thank you that is what I was thinking. I will keep working on this problem
– Arthur Green
Nov 15 at 16:32
4
But if there are $10$ balls, $9$ of which are green, then the probability that the first five are green is $frac12$.
– SmileyCraft
Nov 15 at 16:32
add a comment |
2
No. If there are $10$ balls, $5$ of which are green, the probability that the first five are green is $1big /binom {10}5neq frac 12$.
– lulu
Nov 15 at 16:30
@lulu thank you that is what I was thinking. I will keep working on this problem
– Arthur Green
Nov 15 at 16:32
4
But if there are $10$ balls, $9$ of which are green, then the probability that the first five are green is $frac12$.
– SmileyCraft
Nov 15 at 16:32
2
2
No. If there are $10$ balls, $5$ of which are green, the probability that the first five are green is $1big /binom {10}5neq frac 12$.
– lulu
Nov 15 at 16:30
No. If there are $10$ balls, $5$ of which are green, the probability that the first five are green is $1big /binom {10}5neq frac 12$.
– lulu
Nov 15 at 16:30
@lulu thank you that is what I was thinking. I will keep working on this problem
– Arthur Green
Nov 15 at 16:32
@lulu thank you that is what I was thinking. I will keep working on this problem
– Arthur Green
Nov 15 at 16:32
4
4
But if there are $10$ balls, $9$ of which are green, then the probability that the first five are green is $frac12$.
– SmileyCraft
Nov 15 at 16:32
But if there are $10$ balls, $9$ of which are green, then the probability that the first five are green is $frac12$.
– SmileyCraft
Nov 15 at 16:32
add a comment |
1 Answer
1
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A difference of $1$ ball between green balls and $n$ will yield the minimum $n$ providing it has a whole number solution, so .........
$$frac{binom{n-1}{5}}{binom{n}{5}} = frac{(frac{(n-1)!}{(n-6)!5!})}{(frac{n!}{(n-5)!5!})} = frac{(n-1)!(n-5)!5!}{(n-6)!5!n!} = frac{1}{2}$$
$$frac{(n-1)!(n-5)!}{(n-6)!n!} = frac{1}{2}$$
$$frac{(n-5)}{n} = frac{1}{2}$$
$$2n - 10 = n$$
$$n = 10$$
answered Nov 15 at 18:00
Phil H
3,8782312
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
A difference of $1$ ball between green balls and $n$ will yield the minimum $n$ providing it has a whole number solution, so .........
$$frac{binom{n-1}{5}}{binom{n}{5}} = frac{(frac{(n-1)!}{(n-6)!5!})}{(frac{n!}{(n-5)!5!})} = frac{(n-1)!(n-5)!5!}{(n-6)!5!n!} = frac{1}{2}$$
$$frac{(n-1)!(n-5)!}{(n-6)!n!} = frac{1}{2}$$
$$frac{(n-5)}{n} = frac{1}{2}$$
$$2n - 10 = n$$
$$n = 10$$
answered Nov 15 at 18:00
Phil H
3,8782312
add a comment |
up vote
1
down vote
accepted
A difference of $1$ ball between green balls and $n$ will yield the minimum $n$ providing it has a whole number solution, so .........
$$frac{binom{n-1}{5}}{binom{n}{5}} = frac{(frac{(n-1)!}{(n-6)!5!})}{(frac{n!}{(n-5)!5!})} = frac{(n-1)!(n-5)!5!}{(n-6)!5!n!} = frac{1}{2}$$
$$frac{(n-1)!(n-5)!}{(n-6)!n!} = frac{1}{2}$$
$$frac{(n-5)}{n} = frac{1}{2}$$
$$2n - 10 = n$$
$$n = 10$$
answered Nov 15 at 18:00
Phil H
3,8782312
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
A difference of $1$ ball between green balls and $n$ will yield the minimum $n$ providing it has a whole number solution, so .........
$$frac{binom{n-1}{5}}{binom{n}{5}} = frac{(frac{(n-1)!}{(n-6)!5!})}{(frac{n!}{(n-5)!5!})} = frac{(n-1)!(n-5)!5!}{(n-6)!5!n!} = frac{1}{2}$$
$$frac{(n-1)!(n-5)!}{(n-6)!n!} = frac{1}{2}$$
$$frac{(n-5)}{n} = frac{1}{2}$$
$$2n - 10 = n$$
$$n = 10$$
answered Nov 15 at 18:00
Phil H
3,8782312
A difference of $1$ ball between green balls and $n$ will yield the minimum $n$ providing it has a whole number solution, so .........
$$frac{binom{n-1}{5}}{binom{n}{5}} = frac{(frac{(n-1)!}{(n-6)!5!})}{(frac{n!}{(n-5)!5!})} = frac{(n-1)!(n-5)!5!}{(n-6)!5!n!} = frac{1}{2}$$
$$frac{(n-1)!(n-5)!}{(n-6)!n!} = frac{1}{2}$$
$$frac{(n-5)}{n} = frac{1}{2}$$
$$2n - 10 = n$$
$$n = 10$$
answered Nov 15 at 18:00
Phil H
3,8782312
edited Nov 15 at 18:07
answered Nov 15 at 18:00
Phil H
3,8782312
answered Nov 15 at 18:00
Phil H
3,8782312
answered Nov 15 at 18:00
Phil H
3,8782312
3,8782312
add a comment |
add a comment |
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No. If there are $10$ balls, $5$ of which are green, the probability that the first five are green is $1big /binom {10}5neq frac 12$.
– lulu
Nov 15 at 16:30
@lulu thank you that is what I was thinking. I will keep working on this problem
– Arthur Green
Nov 15 at 16:32
4
But if there are $10$ balls, $9$ of which are green, then the probability that the first five are green is $frac12$.
– SmileyCraft
Nov 15 at 16:32