Finding total number of cases in probability questions
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A pool table has 7 holes through which 5 balls can drop. At each play, each ball is equally likely to go down any of 7 holes. In order to find the probability that each ball passes through distinct holes, I am confused whether the total number of cases is $7^5$ or $5^7$. I always get confused with the power while finding total possibilities.
probability statistics
asked Nov 27 at 7:58
user46697
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A pool table has 7 holes through which 5 balls can drop. At each play, each ball is equally likely to go down any of 7 holes. In order to find the probability that each ball passes through distinct holes, I am confused whether the total number of cases is $7^5$ or $5^7$. I always get confused with the power while finding total possibilities.
probability statistics
asked Nov 27 at 7:58
user46697
193111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A pool table has 7 holes through which 5 balls can drop. At each play, each ball is equally likely to go down any of 7 holes. In order to find the probability that each ball passes through distinct holes, I am confused whether the total number of cases is $7^5$ or $5^7$. I always get confused with the power while finding total possibilities.
probability statistics
asked Nov 27 at 7:58
user46697
193111
A pool table has 7 holes through which 5 balls can drop. At each play, each ball is equally likely to go down any of 7 holes. In order to find the probability that each ball passes through distinct holes, I am confused whether the total number of cases is $7^5$ or $5^7$. I always get confused with the power while finding total possibilities.
probability statistics
probability statistics
asked Nov 27 at 7:58
user46697
193111
asked Nov 27 at 7:58
user46697
193111
asked Nov 27 at 7:58
user46697
193111
asked Nov 27 at 7:58
user46697
193111
asked Nov 27 at 7:58
user46697
193111
193111
add a comment |
add a comment |
2 Answers
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It's useful to think of the "balls going through holes" as a process, and then keep track of the different possibilities so far. First, think of the first ball. It can go through $7$ holes. So we have $7$ cases for the first ball. Then the second ball comes, and it also has $7$ different holes that it can go through. Now we multiply these values and we have $7times 7 = 7^2$ different cases.
Can you see how to continue from here?
answered Nov 27 at 8:02
Matti P.
1,736413
Is it wrong if I think that "a hole can hold any of 5 balls". Then it is 5 possibilities for hole 1. Like wise for the 7 holes wouldnt it be $5^7$.?
– user46697
Nov 27 at 8:33
With that thinking, the process would look like this: First, the first hole chooses one of 5 balls. Then the second hole again chooses one out of 5 balls ... So this approach assumes that every hole takes exactly one ball, and somehow the balls are also able to duplicate! For example, it would be possible that ball #1 goes to all of the holes. Clearly, this is wrong.
– Matti P.
Nov 27 at 8:55
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First you need to choose $5$ holes out of $7$ total in which balls drop. If the balls are indifferent, we have $binom{7}{5}$ cases in total. Otherwise if all are distinct, the number of cases would become $binom{7}{5}5!$ (we assume that the holes have a preserved arrangement)
answered Nov 27 at 8:58
Mostafa Ayaz
13.6k3836
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It's useful to think of the "balls going through holes" as a process, and then keep track of the different possibilities so far. First, think of the first ball. It can go through $7$ holes. So we have $7$ cases for the first ball. Then the second ball comes, and it also has $7$ different holes that it can go through. Now we multiply these values and we have $7times 7 = 7^2$ different cases.
Can you see how to continue from here?
answered Nov 27 at 8:02
Matti P.
1,736413
Is it wrong if I think that "a hole can hold any of 5 balls". Then it is 5 possibilities for hole 1. Like wise for the 7 holes wouldnt it be $5^7$.?
– user46697
Nov 27 at 8:33
With that thinking, the process would look like this: First, the first hole chooses one of 5 balls. Then the second hole again chooses one out of 5 balls ... So this approach assumes that every hole takes exactly one ball, and somehow the balls are also able to duplicate! For example, it would be possible that ball #1 goes to all of the holes. Clearly, this is wrong.
– Matti P.
Nov 27 at 8:55
add a comment |
up vote
2
down vote
It's useful to think of the "balls going through holes" as a process, and then keep track of the different possibilities so far. First, think of the first ball. It can go through $7$ holes. So we have $7$ cases for the first ball. Then the second ball comes, and it also has $7$ different holes that it can go through. Now we multiply these values and we have $7times 7 = 7^2$ different cases.
Can you see how to continue from here?
answered Nov 27 at 8:02
Matti P.
1,736413
Is it wrong if I think that "a hole can hold any of 5 balls". Then it is 5 possibilities for hole 1. Like wise for the 7 holes wouldnt it be $5^7$.?
– user46697
Nov 27 at 8:33
With that thinking, the process would look like this: First, the first hole chooses one of 5 balls. Then the second hole again chooses one out of 5 balls ... So this approach assumes that every hole takes exactly one ball, and somehow the balls are also able to duplicate! For example, it would be possible that ball #1 goes to all of the holes. Clearly, this is wrong.
– Matti P.
Nov 27 at 8:55
add a comment |
up vote
2
down vote
up vote
2
down vote
It's useful to think of the "balls going through holes" as a process, and then keep track of the different possibilities so far. First, think of the first ball. It can go through $7$ holes. So we have $7$ cases for the first ball. Then the second ball comes, and it also has $7$ different holes that it can go through. Now we multiply these values and we have $7times 7 = 7^2$ different cases.
Can you see how to continue from here?
answered Nov 27 at 8:02
Matti P.
1,736413
It's useful to think of the "balls going through holes" as a process, and then keep track of the different possibilities so far. First, think of the first ball. It can go through $7$ holes. So we have $7$ cases for the first ball. Then the second ball comes, and it also has $7$ different holes that it can go through. Now we multiply these values and we have $7times 7 = 7^2$ different cases.
Can you see how to continue from here?
answered Nov 27 at 8:02
Matti P.
1,736413
answered Nov 27 at 8:02
Matti P.
1,736413
answered Nov 27 at 8:02
Matti P.
1,736413
answered Nov 27 at 8:02
Matti P.
1,736413
1,736413
Is it wrong if I think that "a hole can hold any of 5 balls". Then it is 5 possibilities for hole 1. Like wise for the 7 holes wouldnt it be $5^7$.?
– user46697
Nov 27 at 8:33
With that thinking, the process would look like this: First, the first hole chooses one of 5 balls. Then the second hole again chooses one out of 5 balls ... So this approach assumes that every hole takes exactly one ball, and somehow the balls are also able to duplicate! For example, it would be possible that ball #1 goes to all of the holes. Clearly, this is wrong.
– Matti P.
Nov 27 at 8:55
add a comment |
Is it wrong if I think that "a hole can hold any of 5 balls". Then it is 5 possibilities for hole 1. Like wise for the 7 holes wouldnt it be $5^7$.?
– user46697
Nov 27 at 8:33
With that thinking, the process would look like this: First, the first hole chooses one of 5 balls. Then the second hole again chooses one out of 5 balls ... So this approach assumes that every hole takes exactly one ball, and somehow the balls are also able to duplicate! For example, it would be possible that ball #1 goes to all of the holes. Clearly, this is wrong.
– Matti P.
Nov 27 at 8:55
Is it wrong if I think that "a hole can hold any of 5 balls". Then it is 5 possibilities for hole 1. Like wise for the 7 holes wouldnt it be $5^7$.?
– user46697
Nov 27 at 8:33
Is it wrong if I think that "a hole can hold any of 5 balls". Then it is 5 possibilities for hole 1. Like wise for the 7 holes wouldnt it be $5^7$.?
– user46697
Nov 27 at 8:33
With that thinking, the process would look like this: First, the first hole chooses one of 5 balls. Then the second hole again chooses one out of 5 balls ... So this approach assumes that every hole takes exactly one ball, and somehow the balls are also able to duplicate! For example, it would be possible that ball #1 goes to all of the holes. Clearly, this is wrong.
– Matti P.
Nov 27 at 8:55
With that thinking, the process would look like this: First, the first hole chooses one of 5 balls. Then the second hole again chooses one out of 5 balls ... So this approach assumes that every hole takes exactly one ball, and somehow the balls are also able to duplicate! For example, it would be possible that ball #1 goes to all of the holes. Clearly, this is wrong.
– Matti P.
Nov 27 at 8:55
add a comment |
up vote
0
down vote
First you need to choose $5$ holes out of $7$ total in which balls drop. If the balls are indifferent, we have $binom{7}{5}$ cases in total. Otherwise if all are distinct, the number of cases would become $binom{7}{5}5!$ (we assume that the holes have a preserved arrangement)
answered Nov 27 at 8:58
Mostafa Ayaz
13.6k3836
add a comment |
up vote
0
down vote
First you need to choose $5$ holes out of $7$ total in which balls drop. If the balls are indifferent, we have $binom{7}{5}$ cases in total. Otherwise if all are distinct, the number of cases would become $binom{7}{5}5!$ (we assume that the holes have a preserved arrangement)
answered Nov 27 at 8:58
Mostafa Ayaz
13.6k3836
add a comment |
up vote
0
down vote
up vote
0
down vote
First you need to choose $5$ holes out of $7$ total in which balls drop. If the balls are indifferent, we have $binom{7}{5}$ cases in total. Otherwise if all are distinct, the number of cases would become $binom{7}{5}5!$ (we assume that the holes have a preserved arrangement)
answered Nov 27 at 8:58
Mostafa Ayaz
13.6k3836
First you need to choose $5$ holes out of $7$ total in which balls drop. If the balls are indifferent, we have $binom{7}{5}$ cases in total. Otherwise if all are distinct, the number of cases would become $binom{7}{5}5!$ (we assume that the holes have a preserved arrangement)
answered Nov 27 at 8:58
Mostafa Ayaz
13.6k3836
answered Nov 27 at 8:58
Mostafa Ayaz
13.6k3836
answered Nov 27 at 8:58
Mostafa Ayaz
13.6k3836
answered Nov 27 at 8:58
Mostafa Ayaz
13.6k3836
13.6k3836
add a comment |
add a comment |
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