Mixing
Stats and Probability: Three cups and Two different integers in each. Combinations?
$begingroup$
Can’t explain much in the title.
To set the stage:
There are three cups (#1, #2, and #3). In each cup is two cards with different integers. There are 6 different integers that run 1-6. In cup #1 is 1 and 2, in cup #2 is 3 and 4, and so forwardth. The cups are randomized and you have to pull one card from each cup, and then one last card from the last cup you most recently pulled from. What is the probability that one particular string of values would come out?
So:
Pull from “cup #2”: 3
Pull from “cup #1”: 2
Pull from “cup #3”: 6
Make a second draw from “cup #3”: Has to be 5
String is: 3,2,6,5.
-Order matters
-Keep in mind that the cards are replaced after all cards have been pulled.
My example is complicated to me so please ask away for any clarifications.
probability statistics
asked Jan 29 at 21:58
Unhealthy IndividualistUnhealthy Individualist
1
$endgroup$
add a comment |
$begingroup$
Can’t explain much in the title.
To set the stage:
There are three cups (#1, #2, and #3). In each cup is two cards with different integers. There are 6 different integers that run 1-6. In cup #1 is 1 and 2, in cup #2 is 3 and 4, and so forwardth. The cups are randomized and you have to pull one card from each cup, and then one last card from the last cup you most recently pulled from. What is the probability that one particular string of values would come out?
So:
Pull from “cup #2”: 3
Pull from “cup #1”: 2
Pull from “cup #3”: 6
Make a second draw from “cup #3”: Has to be 5
String is: 3,2,6,5.
-Order matters
-Keep in mind that the cards are replaced after all cards have been pulled.
My example is complicated to me so please ask away for any clarifications.
probability statistics
asked Jan 29 at 21:58
Unhealthy IndividualistUnhealthy Individualist
1
$endgroup$
add a comment |
$begingroup$
Can’t explain much in the title.
To set the stage:
There are three cups (#1, #2, and #3). In each cup is two cards with different integers. There are 6 different integers that run 1-6. In cup #1 is 1 and 2, in cup #2 is 3 and 4, and so forwardth. The cups are randomized and you have to pull one card from each cup, and then one last card from the last cup you most recently pulled from. What is the probability that one particular string of values would come out?
So:
Pull from “cup #2”: 3
Pull from “cup #1”: 2
Pull from “cup #3”: 6
Make a second draw from “cup #3”: Has to be 5
String is: 3,2,6,5.
-Order matters
-Keep in mind that the cards are replaced after all cards have been pulled.
My example is complicated to me so please ask away for any clarifications.
probability statistics
asked Jan 29 at 21:58
Unhealthy IndividualistUnhealthy Individualist
1
$endgroup$
Can’t explain much in the title.
To set the stage:
There are three cups (#1, #2, and #3). In each cup is two cards with different integers. There are 6 different integers that run 1-6. In cup #1 is 1 and 2, in cup #2 is 3 and 4, and so forwardth. The cups are randomized and you have to pull one card from each cup, and then one last card from the last cup you most recently pulled from. What is the probability that one particular string of values would come out?
So:
Pull from “cup #2”: 3
Pull from “cup #1”: 2
Pull from “cup #3”: 6
Make a second draw from “cup #3”: Has to be 5
String is: 3,2,6,5.
-Order matters
-Keep in mind that the cards are replaced after all cards have been pulled.
My example is complicated to me so please ask away for any clarifications.
probability statistics
probability statistics
asked Jan 29 at 21:58
Unhealthy IndividualistUnhealthy Individualist
1
asked Jan 29 at 21:58
Unhealthy IndividualistUnhealthy Individualist
1
asked Jan 29 at 21:58
Unhealthy IndividualistUnhealthy Individualist
1
asked Jan 29 at 21:58
Unhealthy IndividualistUnhealthy Individualist
1
asked Jan 29 at 21:58
Unhealthy IndividualistUnhealthy Individualist
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming that the string is valid (i.e., respects the above rules), the probability equals:
$$frac{1}{{6 choose 1}{4 choose 1}{2 choose 1}} = frac{1}{6 cdot 4 cdot 2} = frac{1}{48}$$
This is because you can choose among 6 numbers for the first digit, among 4 numbers for the second, and finally the order of the 2 remaining numbers.
answered Jan 29 at 22:05
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you for guiding me through the the answer!
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:03
$begingroup$
@UnhealthyIndividualist You're welcome. If this or any other answer was of help to you, please don't forget to accept it by ticking the check mark on the left!
$endgroup$
– jvdhooft
Jan 30 at 6:16
add a comment |
$begingroup$
Approach via multiplication principle and counting.
Choose the order of the cups: $3!$ options
Pick a card from the left-most cup: $2$ options
Pick a card from the middle cup: $2$ options
Pick a card from the right-most cup: $2$ options
Pick the remaining card from the right-most cup: $1$ option
Multiplying the number of options gives the number of possible outcomes being $48$. Assuming each outcome is equally likely, the probability of getting a particular possible outcome is simply $1$ divided by the number of outcomes, i.e. $frac{1}{48}$.
answered Jan 29 at 22:04
JMoravitzJMoravitz
48.7k43988
$endgroup$
$begingroup$
Thank you. I forgot about factorials, at least I think that is what this is “!”
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:05
$begingroup$
@UnhealthyIndividualist factorials can also be explained via multiplication principle. To choose the order of the cups, first pick which cup is on the left (3 options), then pick which cup is in the middle (2 options), and the remaining cup goes on the right, for a total of $3cdot 2=6$ options total.
$endgroup$
– JMoravitz
Jan 30 at 2:36
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
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$begingroup$
Assuming that the string is valid (i.e., respects the above rules), the probability equals:
$$frac{1}{{6 choose 1}{4 choose 1}{2 choose 1}} = frac{1}{6 cdot 4 cdot 2} = frac{1}{48}$$
This is because you can choose among 6 numbers for the first digit, among 4 numbers for the second, and finally the order of the 2 remaining numbers.
answered Jan 29 at 22:05
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you for guiding me through the the answer!
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:03
$begingroup$
@UnhealthyIndividualist You're welcome. If this or any other answer was of help to you, please don't forget to accept it by ticking the check mark on the left!
$endgroup$
– jvdhooft
Jan 30 at 6:16
add a comment |
$begingroup$
Assuming that the string is valid (i.e., respects the above rules), the probability equals:
$$frac{1}{{6 choose 1}{4 choose 1}{2 choose 1}} = frac{1}{6 cdot 4 cdot 2} = frac{1}{48}$$
This is because you can choose among 6 numbers for the first digit, among 4 numbers for the second, and finally the order of the 2 remaining numbers.
answered Jan 29 at 22:05
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you for guiding me through the the answer!
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:03
$begingroup$
@UnhealthyIndividualist You're welcome. If this or any other answer was of help to you, please don't forget to accept it by ticking the check mark on the left!
$endgroup$
– jvdhooft
Jan 30 at 6:16
add a comment |
$begingroup$
Assuming that the string is valid (i.e., respects the above rules), the probability equals:
$$frac{1}{{6 choose 1}{4 choose 1}{2 choose 1}} = frac{1}{6 cdot 4 cdot 2} = frac{1}{48}$$
This is because you can choose among 6 numbers for the first digit, among 4 numbers for the second, and finally the order of the 2 remaining numbers.
answered Jan 29 at 22:05
jvdhooftjvdhooft
5,65961641
$endgroup$
Assuming that the string is valid (i.e., respects the above rules), the probability equals:
$$frac{1}{{6 choose 1}{4 choose 1}{2 choose 1}} = frac{1}{6 cdot 4 cdot 2} = frac{1}{48}$$
This is because you can choose among 6 numbers for the first digit, among 4 numbers for the second, and finally the order of the 2 remaining numbers.
answered Jan 29 at 22:05
jvdhooftjvdhooft
5,65961641
answered Jan 29 at 22:05
jvdhooftjvdhooft
5,65961641
answered Jan 29 at 22:05
jvdhooftjvdhooft
5,65961641
answered Jan 29 at 22:05
jvdhooftjvdhooft
5,65961641
5,65961641
$begingroup$
Thank you for guiding me through the the answer!
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:03
$begingroup$
@UnhealthyIndividualist You're welcome. If this or any other answer was of help to you, please don't forget to accept it by ticking the check mark on the left!
$endgroup$
– jvdhooft
Jan 30 at 6:16
add a comment |
$begingroup$
Thank you for guiding me through the the answer!
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:03
$begingroup$
@UnhealthyIndividualist You're welcome. If this or any other answer was of help to you, please don't forget to accept it by ticking the check mark on the left!
$endgroup$
– jvdhooft
Jan 30 at 6:16
$begingroup$
Thank you for guiding me through the the answer!
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:03
$begingroup$
Thank you for guiding me through the the answer!
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:03
$begingroup$
@UnhealthyIndividualist You're welcome. If this or any other answer was of help to you, please don't forget to accept it by ticking the check mark on the left!
$endgroup$
– jvdhooft
Jan 30 at 6:16
$begingroup$
@UnhealthyIndividualist You're welcome. If this or any other answer was of help to you, please don't forget to accept it by ticking the check mark on the left!
$endgroup$
– jvdhooft
Jan 30 at 6:16
add a comment |
$begingroup$
Approach via multiplication principle and counting.
Choose the order of the cups: $3!$ options
Pick a card from the left-most cup: $2$ options
Pick a card from the middle cup: $2$ options
Pick a card from the right-most cup: $2$ options
Pick the remaining card from the right-most cup: $1$ option
Multiplying the number of options gives the number of possible outcomes being $48$. Assuming each outcome is equally likely, the probability of getting a particular possible outcome is simply $1$ divided by the number of outcomes, i.e. $frac{1}{48}$.
answered Jan 29 at 22:04
JMoravitzJMoravitz
48.7k43988
$endgroup$
$begingroup$
Thank you. I forgot about factorials, at least I think that is what this is “!”
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:05
$begingroup$
@UnhealthyIndividualist factorials can also be explained via multiplication principle. To choose the order of the cups, first pick which cup is on the left (3 options), then pick which cup is in the middle (2 options), and the remaining cup goes on the right, for a total of $3cdot 2=6$ options total.
$endgroup$
– JMoravitz
Jan 30 at 2:36
add a comment |
$begingroup$
Approach via multiplication principle and counting.
Choose the order of the cups: $3!$ options
Pick a card from the left-most cup: $2$ options
Pick a card from the middle cup: $2$ options
Pick a card from the right-most cup: $2$ options
Pick the remaining card from the right-most cup: $1$ option
Multiplying the number of options gives the number of possible outcomes being $48$. Assuming each outcome is equally likely, the probability of getting a particular possible outcome is simply $1$ divided by the number of outcomes, i.e. $frac{1}{48}$.
answered Jan 29 at 22:04
JMoravitzJMoravitz
48.7k43988
$endgroup$
$begingroup$
Thank you. I forgot about factorials, at least I think that is what this is “!”
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:05
$begingroup$
@UnhealthyIndividualist factorials can also be explained via multiplication principle. To choose the order of the cups, first pick which cup is on the left (3 options), then pick which cup is in the middle (2 options), and the remaining cup goes on the right, for a total of $3cdot 2=6$ options total.
$endgroup$
– JMoravitz
Jan 30 at 2:36
add a comment |
$begingroup$
Approach via multiplication principle and counting.
Choose the order of the cups: $3!$ options
Pick a card from the left-most cup: $2$ options
Pick a card from the middle cup: $2$ options
Pick a card from the right-most cup: $2$ options
Pick the remaining card from the right-most cup: $1$ option
Multiplying the number of options gives the number of possible outcomes being $48$. Assuming each outcome is equally likely, the probability of getting a particular possible outcome is simply $1$ divided by the number of outcomes, i.e. $frac{1}{48}$.
answered Jan 29 at 22:04
JMoravitzJMoravitz
48.7k43988
$endgroup$
Approach via multiplication principle and counting.
Choose the order of the cups: $3!$ options
Pick a card from the left-most cup: $2$ options
Pick a card from the middle cup: $2$ options
Pick a card from the right-most cup: $2$ options
Pick the remaining card from the right-most cup: $1$ option
Multiplying the number of options gives the number of possible outcomes being $48$. Assuming each outcome is equally likely, the probability of getting a particular possible outcome is simply $1$ divided by the number of outcomes, i.e. $frac{1}{48}$.
answered Jan 29 at 22:04
JMoravitzJMoravitz
48.7k43988
answered Jan 29 at 22:04
JMoravitzJMoravitz
48.7k43988
answered Jan 29 at 22:04
JMoravitzJMoravitz
48.7k43988
answered Jan 29 at 22:04
JMoravitzJMoravitz
48.7k43988
48.7k43988
$begingroup$
Thank you. I forgot about factorials, at least I think that is what this is “!”
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:05
$begingroup$
@UnhealthyIndividualist factorials can also be explained via multiplication principle. To choose the order of the cups, first pick which cup is on the left (3 options), then pick which cup is in the middle (2 options), and the remaining cup goes on the right, for a total of $3cdot 2=6$ options total.
$endgroup$
– JMoravitz
Jan 30 at 2:36
add a comment |
$begingroup$
Thank you. I forgot about factorials, at least I think that is what this is “!”
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:05
$begingroup$
@UnhealthyIndividualist factorials can also be explained via multiplication principle. To choose the order of the cups, first pick which cup is on the left (3 options), then pick which cup is in the middle (2 options), and the remaining cup goes on the right, for a total of $3cdot 2=6$ options total.
$endgroup$
– JMoravitz
Jan 30 at 2:36
$begingroup$
Thank you. I forgot about factorials, at least I think that is what this is “!”
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:05
$begingroup$
Thank you. I forgot about factorials, at least I think that is what this is “!”
$endgroup$
– Unhealthy Individualist
Jan 30 at 2:05
$begingroup$
@UnhealthyIndividualist factorials can also be explained via multiplication principle. To choose the order of the cups, first pick which cup is on the left (3 options), then pick which cup is in the middle (2 options), and the remaining cup goes on the right, for a total of $3cdot 2=6$ options total.
$endgroup$
– JMoravitz
Jan 30 at 2:36
$begingroup$
@UnhealthyIndividualist factorials can also be explained via multiplication principle. To choose the order of the cups, first pick which cup is on the left (3 options), then pick which cup is in the middle (2 options), and the remaining cup goes on the right, for a total of $3cdot 2=6$ options total.
$endgroup$
– JMoravitz
Jan 30 at 2:36
add a comment |
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