Probability lottery question with multiple variables to account for












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Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission. What is the probability of winning the lottery?





I just wanted to check if what I am thinking is correct or not.



Sample size = $80 choose 8$



Winning Events = $11 choose 7$ * $73choose 1$ $~$ $bigg($i.e. $11choose7$ if at least 7 number matches and $73choose 1$ for $8^{th}$ number can be anything $bigg)$



Therefore, the probability is given by:



$$ P = frac{{11 choose 7} {73 choose 1} }{80 choose 8}$$










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission. What is the probability of winning the lottery?





    I just wanted to check if what I am thinking is correct or not.



    Sample size = $80 choose 8$



    Winning Events = $11 choose 7$ * $73choose 1$ $~$ $bigg($i.e. $11choose7$ if at least 7 number matches and $73choose 1$ for $8^{th}$ number can be anything $bigg)$



    Therefore, the probability is given by:



    $$ P = frac{{11 choose 7} {73 choose 1} }{80 choose 8}$$










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission. What is the probability of winning the lottery?





      I just wanted to check if what I am thinking is correct or not.



      Sample size = $80 choose 8$



      Winning Events = $11 choose 7$ * $73choose 1$ $~$ $bigg($i.e. $11choose7$ if at least 7 number matches and $73choose 1$ for $8^{th}$ number can be anything $bigg)$



      Therefore, the probability is given by:



      $$ P = frac{{11 choose 7} {73 choose 1} }{80 choose 8}$$










      share|cite|improve this question









      $endgroup$




      Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission. What is the probability of winning the lottery?





      I just wanted to check if what I am thinking is correct or not.



      Sample size = $80 choose 8$



      Winning Events = $11 choose 7$ * $73choose 1$ $~$ $bigg($i.e. $11choose7$ if at least 7 number matches and $73choose 1$ for $8^{th}$ number can be anything $bigg)$



      Therefore, the probability is given by:



      $$ P = frac{{11 choose 7} {73 choose 1} }{80 choose 8}$$







      probability lotteries






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 6 '18 at 17:11









      zealzeal

      716




      716






















          1 Answer
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          1












          $begingroup$

          Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
            $endgroup$
            – zeal
            Dec 6 '18 at 19:12










          • $begingroup$
            Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:28










          • $begingroup$
            I hope I made clear this :)
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:29












          • $begingroup$
            thank you very much that is clear now.
            $endgroup$
            – zeal
            Dec 7 '18 at 2:09











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          1 Answer
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          1 Answer
          1






          active

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          active

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          active

          oldest

          votes









          1












          $begingroup$

          Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
            $endgroup$
            – zeal
            Dec 6 '18 at 19:12










          • $begingroup$
            Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:28










          • $begingroup$
            I hope I made clear this :)
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:29












          • $begingroup$
            thank you very much that is clear now.
            $endgroup$
            – zeal
            Dec 7 '18 at 2:09
















          1












          $begingroup$

          Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
            $endgroup$
            – zeal
            Dec 6 '18 at 19:12










          • $begingroup$
            Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:28










          • $begingroup$
            I hope I made clear this :)
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:29












          • $begingroup$
            thank you very much that is clear now.
            $endgroup$
            – zeal
            Dec 7 '18 at 2:09














          1












          1








          1





          $begingroup$

          Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$






          share|cite|improve this answer









          $endgroup$



          Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 18:36









          greedoidgreedoid

          41.5k1150102




          41.5k1150102












          • $begingroup$
            I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
            $endgroup$
            – zeal
            Dec 6 '18 at 19:12










          • $begingroup$
            Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:28










          • $begingroup$
            I hope I made clear this :)
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:29












          • $begingroup$
            thank you very much that is clear now.
            $endgroup$
            – zeal
            Dec 7 '18 at 2:09


















          • $begingroup$
            I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
            $endgroup$
            – zeal
            Dec 6 '18 at 19:12










          • $begingroup$
            Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:28










          • $begingroup$
            I hope I made clear this :)
            $endgroup$
            – greedoid
            Dec 6 '18 at 19:29












          • $begingroup$
            thank you very much that is clear now.
            $endgroup$
            – zeal
            Dec 7 '18 at 2:09
















          $begingroup$
          I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
          $endgroup$
          – zeal
          Dec 6 '18 at 19:12




          $begingroup$
          I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
          $endgroup$
          – zeal
          Dec 6 '18 at 19:12












          $begingroup$
          Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
          $endgroup$
          – greedoid
          Dec 6 '18 at 19:28




          $begingroup$
          Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
          $endgroup$
          – greedoid
          Dec 6 '18 at 19:28












          $begingroup$
          I hope I made clear this :)
          $endgroup$
          – greedoid
          Dec 6 '18 at 19:29






          $begingroup$
          I hope I made clear this :)
          $endgroup$
          – greedoid
          Dec 6 '18 at 19:29














          $begingroup$
          thank you very much that is clear now.
          $endgroup$
          – zeal
          Dec 7 '18 at 2:09




          $begingroup$
          thank you very much that is clear now.
          $endgroup$
          – zeal
          Dec 7 '18 at 2:09


















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