Ten targets you need to hit

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You are a good sniper and there are 10 targets you need to hit. The chance of hitting any target is the same.



Strangely, with your accuracy the chance of getting 8 hits is equal to getting 7 hits in 10 hits.



Then




What is the exact chance of hitting all targets?




Note that the chance of getting hit is greater than $0$ and not $100$% either.






probability







share|improve this question























  • Just realised what the question was actually asking so edited my answer to include it :)
    – AHKieran
    32 mins ago














up vote
2
down vote

favorite












You are a good sniper and there are 10 targets you need to hit. The chance of hitting any target is the same.



Strangely, with your accuracy the chance of getting 8 hits is equal to getting 7 hits in 10 hits.



Then




What is the exact chance of hitting all targets?




Note that the chance of getting hit is greater than $0$ and not $100$% either.






probability







share|improve this question























  • Just realised what the question was actually asking so edited my answer to include it :)
    – AHKieran
    32 mins ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











You are a good sniper and there are 10 targets you need to hit. The chance of hitting any target is the same.



Strangely, with your accuracy the chance of getting 8 hits is equal to getting 7 hits in 10 hits.



Then




What is the exact chance of hitting all targets?




Note that the chance of getting hit is greater than $0$ and not $100$% either.






probability







share|improve this question















You are a good sniper and there are 10 targets you need to hit. The chance of hitting any target is the same.



Strangely, with your accuracy the chance of getting 8 hits is equal to getting 7 hits in 10 hits.



Then




What is the exact chance of hitting all targets?




Note that the chance of getting hit is greater than $0$ and not $100$% either.






probability




probability






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 27 mins ago

























asked 1 hour ago









Oray

15.2k435145




15.2k435145











  • Just realised what the question was actually asking so edited my answer to include it :)
    – AHKieran
    32 mins ago
















  • Just realised what the question was actually asking so edited my answer to include it :)
    – AHKieran
    32 mins ago















Just realised what the question was actually asking so edited my answer to include it :)
– AHKieran
32 mins ago




Just realised what the question was actually asking so edited my answer to include it :)
– AHKieran
32 mins ago










5 Answers
5






active

oldest

votes

















up vote
2
down vote













The answer is




$left(frac811right)^10 approx 0.0414$




If $p$ is the probability of hitting a target on any turn then the actual probability of hitting seven targets is




$binom107 p^7 (1-p)^3$




and the chance of hitting eight targets is




$binom108 p^8 (1-p)^2$




That means we want




$binom107(1-p) = binom108p$
$Rightarrow 120(1-p) = 45 p$
$Rightarrow 165p = 120 Rightarrow p = frac811$




and the probability of hitting $10$ is




$p^10 = left(frac811right)^10$







share|improve this answer




















  • I knew I was missing something! Great answer
    – El-Guest
    1 min ago

















up vote
1
down vote













I don't know if this is allowed, but:




Either 0% or 100%.







share|improve this answer




















  • good catch, added a note :)
    – Oray
    57 mins ago

















up vote
1
down vote













If your accuracy is:




75%




Then




Chances are, you'll get 7 or 8 hits, but the chances of it being either are equal.




So




The chance of hitting all the targets is 0.75^10 = 0.0563135147 = ~5.6%







share|improve this answer





























    up vote
    1
    down vote













    The answer is




    $frac12^10 = frac11024$.




    Let




    $p$ be the probability of hitting. By binomial distribution, if the chance of hitting 7 or 8 is identical, then $p^7(1-p)^3 = p^8(1-p)^2 Rightarrow p = 1-p Rightarrow 2p=1 Rightarrow p=frac12$.




    Then the probability of hitting 10 is




    $p^10 = frac12^10 = frac11024$.







    share|improve this answer




















    • Augh, beat me by a minute...!
      – jafe
      20 mins ago










    • i understand how you came to your conclusion but it doesn't sit right in my head. Does this imply that the chance of hitting any number of targets out of the 10 shots is equal?
      – AHKieran
      15 mins ago










    • @jafe close one!! Great minds think alike!
      – El-Guest
      4 mins ago










    • Hexomino has the actual right answer, for what it’s worth — I rushed without considering combinations
      – El-Guest
      2 mins ago

















    up vote
    1
    down vote













    Let's say the chance of hitting one target is p. The chance of hitting 7 or 8 are equal so:




    p^7 * (1-p)^3 * (10 * 9 * 8) = p^8 * (1-p)^2 * (10 *9)




    Since 0 < p < 1, we can




    divide both sides by p^7*(1-p)^2*(10*9).


    (1-p) * 8 = p


    p = 1/9




    So the chance of hitting all targets is




    (1/9)^10







    share|improve this answer






















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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote













      The answer is




      $left(frac811right)^10 approx 0.0414$




      If $p$ is the probability of hitting a target on any turn then the actual probability of hitting seven targets is




      $binom107 p^7 (1-p)^3$




      and the chance of hitting eight targets is




      $binom108 p^8 (1-p)^2$




      That means we want




      $binom107(1-p) = binom108p$
      $Rightarrow 120(1-p) = 45 p$
      $Rightarrow 165p = 120 Rightarrow p = frac811$




      and the probability of hitting $10$ is




      $p^10 = left(frac811right)^10$







      share|improve this answer




















      • I knew I was missing something! Great answer
        – El-Guest
        1 min ago














      up vote
      2
      down vote













      The answer is




      $left(frac811right)^10 approx 0.0414$




      If $p$ is the probability of hitting a target on any turn then the actual probability of hitting seven targets is




      $binom107 p^7 (1-p)^3$




      and the chance of hitting eight targets is




      $binom108 p^8 (1-p)^2$




      That means we want




      $binom107(1-p) = binom108p$
      $Rightarrow 120(1-p) = 45 p$
      $Rightarrow 165p = 120 Rightarrow p = frac811$




      and the probability of hitting $10$ is




      $p^10 = left(frac811right)^10$







      share|improve this answer




















      • I knew I was missing something! Great answer
        – El-Guest
        1 min ago












      up vote
      2
      down vote










      up vote
      2
      down vote









      The answer is




      $left(frac811right)^10 approx 0.0414$




      If $p$ is the probability of hitting a target on any turn then the actual probability of hitting seven targets is




      $binom107 p^7 (1-p)^3$




      and the chance of hitting eight targets is




      $binom108 p^8 (1-p)^2$




      That means we want




      $binom107(1-p) = binom108p$
      $Rightarrow 120(1-p) = 45 p$
      $Rightarrow 165p = 120 Rightarrow p = frac811$




      and the probability of hitting $10$ is




      $p^10 = left(frac811right)^10$







      share|improve this answer












      The answer is




      $left(frac811right)^10 approx 0.0414$




      If $p$ is the probability of hitting a target on any turn then the actual probability of hitting seven targets is




      $binom107 p^7 (1-p)^3$




      and the chance of hitting eight targets is




      $binom108 p^8 (1-p)^2$




      That means we want




      $binom107(1-p) = binom108p$
      $Rightarrow 120(1-p) = 45 p$
      $Rightarrow 165p = 120 Rightarrow p = frac811$




      and the probability of hitting $10$ is




      $p^10 = left(frac811right)^10$








      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 12 mins ago









      hexomino

      30.9k293148




      30.9k293148











      • I knew I was missing something! Great answer
        – El-Guest
        1 min ago
















      • I knew I was missing something! Great answer
        – El-Guest
        1 min ago















      I knew I was missing something! Great answer
      – El-Guest
      1 min ago




      I knew I was missing something! Great answer
      – El-Guest
      1 min ago










      up vote
      1
      down vote













      I don't know if this is allowed, but:




      Either 0% or 100%.







      share|improve this answer




















      • good catch, added a note :)
        – Oray
        57 mins ago














      up vote
      1
      down vote













      I don't know if this is allowed, but:




      Either 0% or 100%.







      share|improve this answer




















      • good catch, added a note :)
        – Oray
        57 mins ago












      up vote
      1
      down vote










      up vote
      1
      down vote









      I don't know if this is allowed, but:




      Either 0% or 100%.







      share|improve this answer












      I don't know if this is allowed, but:




      Either 0% or 100%.








      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 58 mins ago









      Excited Raichu

      60810




      60810











      • good catch, added a note :)
        – Oray
        57 mins ago
















      • good catch, added a note :)
        – Oray
        57 mins ago















      good catch, added a note :)
      – Oray
      57 mins ago




      good catch, added a note :)
      – Oray
      57 mins ago










      up vote
      1
      down vote













      If your accuracy is:




      75%




      Then




      Chances are, you'll get 7 or 8 hits, but the chances of it being either are equal.




      So




      The chance of hitting all the targets is 0.75^10 = 0.0563135147 = ~5.6%







      share|improve this answer


























        up vote
        1
        down vote













        If your accuracy is:




        75%




        Then




        Chances are, you'll get 7 or 8 hits, but the chances of it being either are equal.




        So




        The chance of hitting all the targets is 0.75^10 = 0.0563135147 = ~5.6%







        share|improve this answer
























          up vote
          1
          down vote










          up vote
          1
          down vote









          If your accuracy is:




          75%




          Then




          Chances are, you'll get 7 or 8 hits, but the chances of it being either are equal.




          So




          The chance of hitting all the targets is 0.75^10 = 0.0563135147 = ~5.6%







          share|improve this answer














          If your accuracy is:




          75%




          Then




          Chances are, you'll get 7 or 8 hits, but the chances of it being either are equal.




          So




          The chance of hitting all the targets is 0.75^10 = 0.0563135147 = ~5.6%








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 34 mins ago

























          answered 53 mins ago









          AHKieran

          1,420217




          1,420217




















              up vote
              1
              down vote













              The answer is




              $frac12^10 = frac11024$.




              Let




              $p$ be the probability of hitting. By binomial distribution, if the chance of hitting 7 or 8 is identical, then $p^7(1-p)^3 = p^8(1-p)^2 Rightarrow p = 1-p Rightarrow 2p=1 Rightarrow p=frac12$.




              Then the probability of hitting 10 is




              $p^10 = frac12^10 = frac11024$.







              share|improve this answer




















              • Augh, beat me by a minute...!
                – jafe
                20 mins ago










              • i understand how you came to your conclusion but it doesn't sit right in my head. Does this imply that the chance of hitting any number of targets out of the 10 shots is equal?
                – AHKieran
                15 mins ago










              • @jafe close one!! Great minds think alike!
                – El-Guest
                4 mins ago










              • Hexomino has the actual right answer, for what it’s worth — I rushed without considering combinations
                – El-Guest
                2 mins ago














              up vote
              1
              down vote













              The answer is




              $frac12^10 = frac11024$.




              Let




              $p$ be the probability of hitting. By binomial distribution, if the chance of hitting 7 or 8 is identical, then $p^7(1-p)^3 = p^8(1-p)^2 Rightarrow p = 1-p Rightarrow 2p=1 Rightarrow p=frac12$.




              Then the probability of hitting 10 is




              $p^10 = frac12^10 = frac11024$.







              share|improve this answer




















              • Augh, beat me by a minute...!
                – jafe
                20 mins ago










              • i understand how you came to your conclusion but it doesn't sit right in my head. Does this imply that the chance of hitting any number of targets out of the 10 shots is equal?
                – AHKieran
                15 mins ago










              • @jafe close one!! Great minds think alike!
                – El-Guest
                4 mins ago










              • Hexomino has the actual right answer, for what it’s worth — I rushed without considering combinations
                – El-Guest
                2 mins ago












              up vote
              1
              down vote










              up vote
              1
              down vote









              The answer is




              $frac12^10 = frac11024$.




              Let




              $p$ be the probability of hitting. By binomial distribution, if the chance of hitting 7 or 8 is identical, then $p^7(1-p)^3 = p^8(1-p)^2 Rightarrow p = 1-p Rightarrow 2p=1 Rightarrow p=frac12$.




              Then the probability of hitting 10 is




              $p^10 = frac12^10 = frac11024$.







              share|improve this answer












              The answer is




              $frac12^10 = frac11024$.




              Let




              $p$ be the probability of hitting. By binomial distribution, if the chance of hitting 7 or 8 is identical, then $p^7(1-p)^3 = p^8(1-p)^2 Rightarrow p = 1-p Rightarrow 2p=1 Rightarrow p=frac12$.




              Then the probability of hitting 10 is




              $p^10 = frac12^10 = frac11024$.








              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 24 mins ago









              El-Guest

              13.5k2965




              13.5k2965











              • Augh, beat me by a minute...!
                – jafe
                20 mins ago










              • i understand how you came to your conclusion but it doesn't sit right in my head. Does this imply that the chance of hitting any number of targets out of the 10 shots is equal?
                – AHKieran
                15 mins ago










              • @jafe close one!! Great minds think alike!
                – El-Guest
                4 mins ago










              • Hexomino has the actual right answer, for what it’s worth — I rushed without considering combinations
                – El-Guest
                2 mins ago
















              • Augh, beat me by a minute...!
                – jafe
                20 mins ago










              • i understand how you came to your conclusion but it doesn't sit right in my head. Does this imply that the chance of hitting any number of targets out of the 10 shots is equal?
                – AHKieran
                15 mins ago










              • @jafe close one!! Great minds think alike!
                – El-Guest
                4 mins ago










              • Hexomino has the actual right answer, for what it’s worth — I rushed without considering combinations
                – El-Guest
                2 mins ago















              Augh, beat me by a minute...!
              – jafe
              20 mins ago




              Augh, beat me by a minute...!
              – jafe
              20 mins ago












              i understand how you came to your conclusion but it doesn't sit right in my head. Does this imply that the chance of hitting any number of targets out of the 10 shots is equal?
              – AHKieran
              15 mins ago




              i understand how you came to your conclusion but it doesn't sit right in my head. Does this imply that the chance of hitting any number of targets out of the 10 shots is equal?
              – AHKieran
              15 mins ago












              @jafe close one!! Great minds think alike!
              – El-Guest
              4 mins ago




              @jafe close one!! Great minds think alike!
              – El-Guest
              4 mins ago












              Hexomino has the actual right answer, for what it’s worth — I rushed without considering combinations
              – El-Guest
              2 mins ago




              Hexomino has the actual right answer, for what it’s worth — I rushed without considering combinations
              – El-Guest
              2 mins ago










              up vote
              1
              down vote













              Let's say the chance of hitting one target is p. The chance of hitting 7 or 8 are equal so:




              p^7 * (1-p)^3 * (10 * 9 * 8) = p^8 * (1-p)^2 * (10 *9)




              Since 0 < p < 1, we can




              divide both sides by p^7*(1-p)^2*(10*9).


              (1-p) * 8 = p


              p = 1/9




              So the chance of hitting all targets is




              (1/9)^10







              share|improve this answer


























                up vote
                1
                down vote













                Let's say the chance of hitting one target is p. The chance of hitting 7 or 8 are equal so:




                p^7 * (1-p)^3 * (10 * 9 * 8) = p^8 * (1-p)^2 * (10 *9)




                Since 0 < p < 1, we can




                divide both sides by p^7*(1-p)^2*(10*9).


                (1-p) * 8 = p


                p = 1/9




                So the chance of hitting all targets is




                (1/9)^10







                share|improve this answer
























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Let's say the chance of hitting one target is p. The chance of hitting 7 or 8 are equal so:




                  p^7 * (1-p)^3 * (10 * 9 * 8) = p^8 * (1-p)^2 * (10 *9)




                  Since 0 < p < 1, we can




                  divide both sides by p^7*(1-p)^2*(10*9).


                  (1-p) * 8 = p


                  p = 1/9




                  So the chance of hitting all targets is




                  (1/9)^10







                  share|improve this answer














                  Let's say the chance of hitting one target is p. The chance of hitting 7 or 8 are equal so:




                  p^7 * (1-p)^3 * (10 * 9 * 8) = p^8 * (1-p)^2 * (10 *9)




                  Since 0 < p < 1, we can




                  divide both sides by p^7*(1-p)^2*(10*9).


                  (1-p) * 8 = p


                  p = 1/9




                  So the chance of hitting all targets is




                  (1/9)^10








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 8 mins ago

























                  answered 21 mins ago









                  jafe

                  6,7441682




                  6,7441682



























                       

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