Mixing
Probability for red and green apples in a bag
0
$begingroup$
"There are twelve red apples and seven green apples in a bag. What is the probability of picking up one red apple and one green apple at once?"
This is a multiple choice question, the choices are:
- A) $frac{21}{64}$
- B) $frac{23}{64}$
- C) $frac{25}{64}$
- D) $frac{22}{64}$
The "at once" is throwing me off but I interpreted it as picking red then green or green then red without replacement, which gave me $frac{28}{57}$. I have no idea how all the answers have $64$ as the denominator, is there something obvious I'm missing?
probability
edited Jan 19 at 1:33
EuxhenH
482210
asked Jan 19 at 0:35
togekisstogekiss
6
$endgroup$
add a comment |
0
$begingroup$
"There are twelve red apples and seven green apples in a bag. What is the probability of picking up one red apple and one green apple at once?"
This is a multiple choice question, the choices are:
- A) $frac{21}{64}$
- B) $frac{23}{64}$
- C) $frac{25}{64}$
- D) $frac{22}{64}$
The "at once" is throwing me off but I interpreted it as picking red then green or green then red without replacement, which gave me $frac{28}{57}$. I have no idea how all the answers have $64$ as the denominator, is there something obvious I'm missing?
probability
edited Jan 19 at 1:33
EuxhenH
482210
asked Jan 19 at 0:35
togekisstogekiss
6
$endgroup$
$begingroup$
I would have interpreted this as meaning, "You pull out two apples at once. What is the probability that one is read and one green?" but that gives me $7cdot12/{19choose2}=84/171$
$endgroup$
– saulspatz
Jan 19 at 0:48
$begingroup$
Welcome to MathSE. I agree with your calculation. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 19 at 0:49
3
$begingroup$
I agree with the interpretation that leads to the answer of $frac{28}{57}=frac{7cdot 12}{binom{19}{2}}$. I see no possible interpretation which leads to a denominator being $64$ or any multiple of $64$ except for if there happen to also be yellow apples in the bag as well that you forgot to mention or the numbers of apples you write are incorrect. As far as we can tell you did nothing wrong and the answer key is what is incorrect. If it really is that the answer key is incorrect, then make note of it and bring it up with the teacher.
$endgroup$
– JMoravitz
Jan 19 at 0:55
add a comment |
0
0
0
$begingroup$
"There are twelve red apples and seven green apples in a bag. What is the probability of picking up one red apple and one green apple at once?"
This is a multiple choice question, the choices are:
- A) $frac{21}{64}$
- B) $frac{23}{64}$
- C) $frac{25}{64}$
- D) $frac{22}{64}$
The "at once" is throwing me off but I interpreted it as picking red then green or green then red without replacement, which gave me $frac{28}{57}$. I have no idea how all the answers have $64$ as the denominator, is there something obvious I'm missing?
probability
edited Jan 19 at 1:33
EuxhenH
482210
asked Jan 19 at 0:35
togekisstogekiss
6
$endgroup$
"There are twelve red apples and seven green apples in a bag. What is the probability of picking up one red apple and one green apple at once?"
This is a multiple choice question, the choices are:
- A) $frac{21}{64}$
- B) $frac{23}{64}$
- C) $frac{25}{64}$
- D) $frac{22}{64}$
The "at once" is throwing me off but I interpreted it as picking red then green or green then red without replacement, which gave me $frac{28}{57}$. I have no idea how all the answers have $64$ as the denominator, is there something obvious I'm missing?
probability
probability
edited Jan 19 at 1:33
EuxhenH
482210
asked Jan 19 at 0:35
togekisstogekiss
6
edited Jan 19 at 1:33
EuxhenH
482210
asked Jan 19 at 0:35
togekisstogekiss
6
edited Jan 19 at 1:33
EuxhenH
482210
edited Jan 19 at 1:33
EuxhenH
482210
edited Jan 19 at 1:33
EuxhenH
482210
482210
asked Jan 19 at 0:35
togekisstogekiss
6
asked Jan 19 at 0:35
togekisstogekiss
6
asked Jan 19 at 0:35
togekisstogekiss
6
6
$begingroup$
I would have interpreted this as meaning, "You pull out two apples at once. What is the probability that one is read and one green?" but that gives me $7cdot12/{19choose2}=84/171$
$endgroup$
– saulspatz
Jan 19 at 0:48
$begingroup$
Welcome to MathSE. I agree with your calculation. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 19 at 0:49
3
$begingroup$
I agree with the interpretation that leads to the answer of $frac{28}{57}=frac{7cdot 12}{binom{19}{2}}$. I see no possible interpretation which leads to a denominator being $64$ or any multiple of $64$ except for if there happen to also be yellow apples in the bag as well that you forgot to mention or the numbers of apples you write are incorrect. As far as we can tell you did nothing wrong and the answer key is what is incorrect. If it really is that the answer key is incorrect, then make note of it and bring it up with the teacher.
$endgroup$
– JMoravitz
Jan 19 at 0:55
add a comment |
$begingroup$
I would have interpreted this as meaning, "You pull out two apples at once. What is the probability that one is read and one green?" but that gives me $7cdot12/{19choose2}=84/171$
$endgroup$
– saulspatz
Jan 19 at 0:48
$begingroup$
Welcome to MathSE. I agree with your calculation. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 19 at 0:49
3
$begingroup$
I agree with the interpretation that leads to the answer of $frac{28}{57}=frac{7cdot 12}{binom{19}{2}}$. I see no possible interpretation which leads to a denominator being $64$ or any multiple of $64$ except for if there happen to also be yellow apples in the bag as well that you forgot to mention or the numbers of apples you write are incorrect. As far as we can tell you did nothing wrong and the answer key is what is incorrect. If it really is that the answer key is incorrect, then make note of it and bring it up with the teacher.
$endgroup$
– JMoravitz
Jan 19 at 0:55
$begingroup$
I would have interpreted this as meaning, "You pull out two apples at once. What is the probability that one is read and one green?" but that gives me $7cdot12/{19choose2}=84/171$
$endgroup$
– saulspatz
Jan 19 at 0:48
$begingroup$
I would have interpreted this as meaning, "You pull out two apples at once. What is the probability that one is read and one green?" but that gives me $7cdot12/{19choose2}=84/171$
$endgroup$
– saulspatz
Jan 19 at 0:48
$begingroup$
Welcome to MathSE. I agree with your calculation. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 19 at 0:49
$begingroup$
Welcome to MathSE. I agree with your calculation. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 19 at 0:49
3
3
$begingroup$
I agree with the interpretation that leads to the answer of $frac{28}{57}=frac{7cdot 12}{binom{19}{2}}$. I see no possible interpretation which leads to a denominator being $64$ or any multiple of $64$ except for if there happen to also be yellow apples in the bag as well that you forgot to mention or the numbers of apples you write are incorrect. As far as we can tell you did nothing wrong and the answer key is what is incorrect. If it really is that the answer key is incorrect, then make note of it and bring it up with the teacher.
$endgroup$
– JMoravitz
Jan 19 at 0:55
$begingroup$
I agree with the interpretation that leads to the answer of $frac{28}{57}=frac{7cdot 12}{binom{19}{2}}$. I see no possible interpretation which leads to a denominator being $64$ or any multiple of $64$ except for if there happen to also be yellow apples in the bag as well that you forgot to mention or the numbers of apples you write are incorrect. As far as we can tell you did nothing wrong and the answer key is what is incorrect. If it really is that the answer key is incorrect, then make note of it and bring it up with the teacher.
$endgroup$
– JMoravitz
Jan 19 at 0:55
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
I would argue that the answer $frac{28}{57}$ is correct.
The argument:
Number of ways of choosing a red apple and a green apple, = $7.12$
Total number of ways of choosing two objects out of $19$ = $19 choose{2}$
Therefore, the answer comes out as, $frac{7.12}{19 choose 2} = frac{28}{57}$
answered Jan 19 at 1:07
The Jade EmperorThe Jade Emperor
908
$endgroup$
1
$begingroup$
The division by $2$ is not necessary. To see this, consider the problem where it was instead $7$ red apples and $11$ green apples. If you were to divide by $2$ in that case then you would have a non-integer number of arrangements, a clear impossibility.
$endgroup$
– JMoravitz
Jan 19 at 1:39
1
$begingroup$
If that argument isn't convincing enough, consider the problem where we have one red apple and one green apple in a bag and we choose two of them. We ask what the probability is that we select both a red and a green apple. Without even calculating you should be able to tell that the answer should be $1$, but following your proposed formula you'd only get an answer of $frac{1}{2}$.
$endgroup$
– JMoravitz
Jan 19 at 2:51
$begingroup$
True. I will correct it. Thanks!!
$endgroup$
– The Jade Emperor
Jan 19 at 3:07
add a comment |
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0
$begingroup$
I would argue that the answer $frac{28}{57}$ is correct.
The argument:
Number of ways of choosing a red apple and a green apple, = $7.12$
Total number of ways of choosing two objects out of $19$ = $19 choose{2}$
Therefore, the answer comes out as, $frac{7.12}{19 choose 2} = frac{28}{57}$
answered Jan 19 at 1:07
The Jade EmperorThe Jade Emperor
908
$endgroup$
1
$begingroup$
The division by $2$ is not necessary. To see this, consider the problem where it was instead $7$ red apples and $11$ green apples. If you were to divide by $2$ in that case then you would have a non-integer number of arrangements, a clear impossibility.
$endgroup$
– JMoravitz
Jan 19 at 1:39
1
$begingroup$
If that argument isn't convincing enough, consider the problem where we have one red apple and one green apple in a bag and we choose two of them. We ask what the probability is that we select both a red and a green apple. Without even calculating you should be able to tell that the answer should be $1$, but following your proposed formula you'd only get an answer of $frac{1}{2}$.
$endgroup$
– JMoravitz
Jan 19 at 2:51
$begingroup$
True. I will correct it. Thanks!!
$endgroup$
– The Jade Emperor
Jan 19 at 3:07
add a comment |
0
$begingroup$
I would argue that the answer $frac{28}{57}$ is correct.
The argument:
Number of ways of choosing a red apple and a green apple, = $7.12$
Total number of ways of choosing two objects out of $19$ = $19 choose{2}$
Therefore, the answer comes out as, $frac{7.12}{19 choose 2} = frac{28}{57}$
answered Jan 19 at 1:07
The Jade EmperorThe Jade Emperor
908
$endgroup$
1
$begingroup$
The division by $2$ is not necessary. To see this, consider the problem where it was instead $7$ red apples and $11$ green apples. If you were to divide by $2$ in that case then you would have a non-integer number of arrangements, a clear impossibility.
$endgroup$
– JMoravitz
Jan 19 at 1:39
1
$begingroup$
If that argument isn't convincing enough, consider the problem where we have one red apple and one green apple in a bag and we choose two of them. We ask what the probability is that we select both a red and a green apple. Without even calculating you should be able to tell that the answer should be $1$, but following your proposed formula you'd only get an answer of $frac{1}{2}$.
$endgroup$
– JMoravitz
Jan 19 at 2:51
$begingroup$
True. I will correct it. Thanks!!
$endgroup$
– The Jade Emperor
Jan 19 at 3:07
add a comment |
0
0
0
$begingroup$
I would argue that the answer $frac{28}{57}$ is correct.
The argument:
Number of ways of choosing a red apple and a green apple, = $7.12$
Total number of ways of choosing two objects out of $19$ = $19 choose{2}$
Therefore, the answer comes out as, $frac{7.12}{19 choose 2} = frac{28}{57}$
answered Jan 19 at 1:07
The Jade EmperorThe Jade Emperor
908
$endgroup$
I would argue that the answer $frac{28}{57}$ is correct.
The argument:
Number of ways of choosing a red apple and a green apple, = $7.12$
Total number of ways of choosing two objects out of $19$ = $19 choose{2}$
Therefore, the answer comes out as, $frac{7.12}{19 choose 2} = frac{28}{57}$
answered Jan 19 at 1:07
The Jade EmperorThe Jade Emperor
908
edited Jan 19 at 3:08
answered Jan 19 at 1:07
The Jade EmperorThe Jade Emperor
908
answered Jan 19 at 1:07
The Jade EmperorThe Jade Emperor
908
answered Jan 19 at 1:07
The Jade EmperorThe Jade Emperor
908
908
1
$begingroup$
The division by $2$ is not necessary. To see this, consider the problem where it was instead $7$ red apples and $11$ green apples. If you were to divide by $2$ in that case then you would have a non-integer number of arrangements, a clear impossibility.
$endgroup$
– JMoravitz
Jan 19 at 1:39
1
$begingroup$
If that argument isn't convincing enough, consider the problem where we have one red apple and one green apple in a bag and we choose two of them. We ask what the probability is that we select both a red and a green apple. Without even calculating you should be able to tell that the answer should be $1$, but following your proposed formula you'd only get an answer of $frac{1}{2}$.
$endgroup$
– JMoravitz
Jan 19 at 2:51
$begingroup$
True. I will correct it. Thanks!!
$endgroup$
– The Jade Emperor
Jan 19 at 3:07
add a comment |
1
$begingroup$
The division by $2$ is not necessary. To see this, consider the problem where it was instead $7$ red apples and $11$ green apples. If you were to divide by $2$ in that case then you would have a non-integer number of arrangements, a clear impossibility.
$endgroup$
– JMoravitz
Jan 19 at 1:39
1
$begingroup$
If that argument isn't convincing enough, consider the problem where we have one red apple and one green apple in a bag and we choose two of them. We ask what the probability is that we select both a red and a green apple. Without even calculating you should be able to tell that the answer should be $1$, but following your proposed formula you'd only get an answer of $frac{1}{2}$.
$endgroup$
– JMoravitz
Jan 19 at 2:51
$begingroup$
True. I will correct it. Thanks!!
$endgroup$
– The Jade Emperor
Jan 19 at 3:07
1
1
$begingroup$
The division by $2$ is not necessary. To see this, consider the problem where it was instead $7$ red apples and $11$ green apples. If you were to divide by $2$ in that case then you would have a non-integer number of arrangements, a clear impossibility.
$endgroup$
– JMoravitz
Jan 19 at 1:39
$begingroup$
The division by $2$ is not necessary. To see this, consider the problem where it was instead $7$ red apples and $11$ green apples. If you were to divide by $2$ in that case then you would have a non-integer number of arrangements, a clear impossibility.
$endgroup$
– JMoravitz
Jan 19 at 1:39
1
1
$begingroup$
If that argument isn't convincing enough, consider the problem where we have one red apple and one green apple in a bag and we choose two of them. We ask what the probability is that we select both a red and a green apple. Without even calculating you should be able to tell that the answer should be $1$, but following your proposed formula you'd only get an answer of $frac{1}{2}$.
$endgroup$
– JMoravitz
Jan 19 at 2:51
$begingroup$
If that argument isn't convincing enough, consider the problem where we have one red apple and one green apple in a bag and we choose two of them. We ask what the probability is that we select both a red and a green apple. Without even calculating you should be able to tell that the answer should be $1$, but following your proposed formula you'd only get an answer of $frac{1}{2}$.
$endgroup$
– JMoravitz
Jan 19 at 2:51
$begingroup$
True. I will correct it. Thanks!!
$endgroup$
– The Jade Emperor
Jan 19 at 3:07
$begingroup$
True. I will correct it. Thanks!!
$endgroup$
– The Jade Emperor
Jan 19 at 3:07
add a comment |
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$begingroup$
I would have interpreted this as meaning, "You pull out two apples at once. What is the probability that one is read and one green?" but that gives me $7cdot12/{19choose2}=84/171$
$endgroup$
– saulspatz
Jan 19 at 0:48
$begingroup$
Welcome to MathSE. I agree with your calculation. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 19 at 0:49
3
$begingroup$
I agree with the interpretation that leads to the answer of $frac{28}{57}=frac{7cdot 12}{binom{19}{2}}$. I see no possible interpretation which leads to a denominator being $64$ or any multiple of $64$ except for if there happen to also be yellow apples in the bag as well that you forgot to mention or the numbers of apples you write are incorrect. As far as we can tell you did nothing wrong and the answer key is what is incorrect. If it really is that the answer key is incorrect, then make note of it and bring it up with the teacher.
$endgroup$
– JMoravitz
Jan 19 at 0:55