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.










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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.










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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.










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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






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      asked Nov 27 at 7:58









      user46697

      193111




      193111






















          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?






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          • 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



















          0














          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)






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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            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?






            share|cite|improve this answer





















            • 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
















            2














            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?






            share|cite|improve this answer





















            • 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














            2












            2








            2






            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?






            share|cite|improve this answer












            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?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 27 at 8:02









            Matti P.

            1,738413




            1,738413












            • 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










            • 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











            0














            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)






            share|cite|improve this answer


























              0














              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)






              share|cite|improve this answer
























                0












                0








                0






                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)






                share|cite|improve this answer












                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)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 at 8:58









                Mostafa Ayaz

                13.7k3836




                13.7k3836






























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