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Complicated probability question - Can you figure out the composition of the marbles in this urn?
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There is an urn filled with millions of fancy marbles. Anytime a marble is drawn from the urn, an identical version is immediately put back into the urn by the manufacturing technology.
The marbles come in 10 different colors, each with its own price. The prices range (uniformly) from 8 dollars to 12 dollars, and the average price is 10 dollars. If customers wish to purchase, they must buy six marbles - no more and no less.
You do not know how likely it is to draw a particular color marble from the urn, but it is not uniform. For example, red marbles are more prevalent than other colors. (They are also the most expensive at 12 dollars.)
By way of customer data you have access to the following information:
Of the marble sets which contain six different (unique) colors and add up to less than 60 dollars, approximately 23 percent of those sets contain a red marble.
Can you calculate the probability of drawing a red marble from the urn on a single draw?
probability probability-distributions binomial-distribution
asked Aug 6 at 14:18
Miles Coltrane
113
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up vote
2
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favorite
There is an urn filled with millions of fancy marbles. Anytime a marble is drawn from the urn, an identical version is immediately put back into the urn by the manufacturing technology.
The marbles come in 10 different colors, each with its own price. The prices range (uniformly) from 8 dollars to 12 dollars, and the average price is 10 dollars. If customers wish to purchase, they must buy six marbles - no more and no less.
You do not know how likely it is to draw a particular color marble from the urn, but it is not uniform. For example, red marbles are more prevalent than other colors. (They are also the most expensive at 12 dollars.)
By way of customer data you have access to the following information:
Of the marble sets which contain six different (unique) colors and add up to less than 60 dollars, approximately 23 percent of those sets contain a red marble.
Can you calculate the probability of drawing a red marble from the urn on a single draw?
probability probability-distributions binomial-distribution
asked Aug 6 at 14:18
Miles Coltrane
113
1
Assuming the costs of the colors are $8,8,9,9,10,10,11,11,12,12$, there are 88 sets of 6 colors which add up to less than $60, so how can exactly 23% of them have a red marble?
– Akababa
Aug 6 at 15:23
You are correct - I rounded up. Assume approximately 23%. I will edit now.
– Miles Coltrane
Aug 6 at 15:27
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
There is an urn filled with millions of fancy marbles. Anytime a marble is drawn from the urn, an identical version is immediately put back into the urn by the manufacturing technology.
The marbles come in 10 different colors, each with its own price. The prices range (uniformly) from 8 dollars to 12 dollars, and the average price is 10 dollars. If customers wish to purchase, they must buy six marbles - no more and no less.
You do not know how likely it is to draw a particular color marble from the urn, but it is not uniform. For example, red marbles are more prevalent than other colors. (They are also the most expensive at 12 dollars.)
By way of customer data you have access to the following information:
Of the marble sets which contain six different (unique) colors and add up to less than 60 dollars, approximately 23 percent of those sets contain a red marble.
Can you calculate the probability of drawing a red marble from the urn on a single draw?
probability probability-distributions binomial-distribution
asked Aug 6 at 14:18
Miles Coltrane
113
There is an urn filled with millions of fancy marbles. Anytime a marble is drawn from the urn, an identical version is immediately put back into the urn by the manufacturing technology.
The marbles come in 10 different colors, each with its own price. The prices range (uniformly) from 8 dollars to 12 dollars, and the average price is 10 dollars. If customers wish to purchase, they must buy six marbles - no more and no less.
You do not know how likely it is to draw a particular color marble from the urn, but it is not uniform. For example, red marbles are more prevalent than other colors. (They are also the most expensive at 12 dollars.)
By way of customer data you have access to the following information:
Of the marble sets which contain six different (unique) colors and add up to less than 60 dollars, approximately 23 percent of those sets contain a red marble.
Can you calculate the probability of drawing a red marble from the urn on a single draw?
probability probability-distributions binomial-distribution
asked Aug 6 at 14:18
Miles Coltrane
113
edited Aug 6 at 19:03
asked Aug 6 at 14:18
Miles Coltrane
113
asked Aug 6 at 14:18
Miles Coltrane
113
asked Aug 6 at 14:18
Miles Coltrane
113
113
1
Assuming the costs of the colors are $8,8,9,9,10,10,11,11,12,12$, there are 88 sets of 6 colors which add up to less than $60, so how can exactly 23% of them have a red marble?
– Akababa
Aug 6 at 15:23
You are correct - I rounded up. Assume approximately 23%. I will edit now.
– Miles Coltrane
Aug 6 at 15:27
add a comment |Â
1
Assuming the costs of the colors are $8,8,9,9,10,10,11,11,12,12$, there are 88 sets of 6 colors which add up to less than $60, so how can exactly 23% of them have a red marble?
– Akababa
Aug 6 at 15:23
You are correct - I rounded up. Assume approximately 23%. I will edit now.
– Miles Coltrane
Aug 6 at 15:27
1
1
Assuming the costs of the colors are $8,8,9,9,10,10,11,11,12,12$, there are 88 sets of 6 colors which add up to less than $60, so how can exactly 23% of them have a red marble?
– Akababa
Aug 6 at 15:23
Assuming the costs of the colors are $8,8,9,9,10,10,11,11,12,12$, there are 88 sets of 6 colors which add up to less than $60, so how can exactly 23% of them have a red marble?
– Akababa
Aug 6 at 15:23
You are correct - I rounded up. Assume approximately 23%. I will edit now.
– Miles Coltrane
Aug 6 at 15:27
You are correct - I rounded up. Assume approximately 23%. I will edit now.
– Miles Coltrane
Aug 6 at 15:27
add a comment |Â
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1
Assuming the costs of the colors are $8,8,9,9,10,10,11,11,12,12$, there are 88 sets of 6 colors which add up to less than $60, so how can exactly 23% of them have a red marble?
– Akababa
Aug 6 at 15:23
You are correct - I rounded up. Assume approximately 23%. I will edit now.
– Miles Coltrane
Aug 6 at 15:27