Prob choosing 1 out of 2 from 11












-1












$begingroup$


Suppose we have 11 items, 2 of which are "good". Someone wins a prize if in 1 turn they pick either of the 2 good items.



Whats the probability the person wins?



I got 3.6% using combinations but not sure if its correct



I want to be able to extend this out into say, if there was 3 items out of the 11 which were good, and the contestant had two goes to choose an item.






probability







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    That is far too low. If you just choose $1$ the probability that it is good is $frac 2{11}=18.18%$.
    $endgroup$
    – lulu
    Dec 13 '18 at 15:58






  • 1




    $begingroup$
    Your formulation is rather vague: "has the chance to win a prize if.."?... So picking the right one only gives a chance on winning a price? Secondly, what is a "turn" here?
    $endgroup$
    – drhab
    Dec 13 '18 at 15:59








  • 1




    $begingroup$
    Please write more clearly. What does "choose 2 from 3 out of 11 where 3 are good" mean?
    $endgroup$
    – lulu
    Dec 13 '18 at 15:59








  • 1




    $begingroup$
    If you are saying: there are three good ones and eight bad ones and you choose two, then the probability that both are bad is $frac 8{11}times frac 7{10}=frac {28}{55}approx 51%$. But is that what you meant?
    $endgroup$
    – lulu
    Dec 13 '18 at 16:01








  • 2




    $begingroup$
    Oh, you need both to be good? Ok, then it is $frac 3{11}times frac 2{10}=frac 3{55}$.
    $endgroup$
    – lulu
    Dec 13 '18 at 16:02
















-1












$begingroup$


Suppose we have 11 items, 2 of which are "good". Someone wins a prize if in 1 turn they pick either of the 2 good items.



Whats the probability the person wins?



I got 3.6% using combinations but not sure if its correct



I want to be able to extend this out into say, if there was 3 items out of the 11 which were good, and the contestant had two goes to choose an item.






probability







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    That is far too low. If you just choose $1$ the probability that it is good is $frac 2{11}=18.18%$.
    $endgroup$
    – lulu
    Dec 13 '18 at 15:58






  • 1




    $begingroup$
    Your formulation is rather vague: "has the chance to win a prize if.."?... So picking the right one only gives a chance on winning a price? Secondly, what is a "turn" here?
    $endgroup$
    – drhab
    Dec 13 '18 at 15:59








  • 1




    $begingroup$
    Please write more clearly. What does "choose 2 from 3 out of 11 where 3 are good" mean?
    $endgroup$
    – lulu
    Dec 13 '18 at 15:59








  • 1




    $begingroup$
    If you are saying: there are three good ones and eight bad ones and you choose two, then the probability that both are bad is $frac 8{11}times frac 7{10}=frac {28}{55}approx 51%$. But is that what you meant?
    $endgroup$
    – lulu
    Dec 13 '18 at 16:01








  • 2




    $begingroup$
    Oh, you need both to be good? Ok, then it is $frac 3{11}times frac 2{10}=frac 3{55}$.
    $endgroup$
    – lulu
    Dec 13 '18 at 16:02














-1












-1








-1





$begingroup$


Suppose we have 11 items, 2 of which are "good". Someone wins a prize if in 1 turn they pick either of the 2 good items.



Whats the probability the person wins?



I got 3.6% using combinations but not sure if its correct



I want to be able to extend this out into say, if there was 3 items out of the 11 which were good, and the contestant had two goes to choose an item.






probability







share|cite|improve this question











$endgroup$




Suppose we have 11 items, 2 of which are "good". Someone wins a prize if in 1 turn they pick either of the 2 good items.



Whats the probability the person wins?



I got 3.6% using combinations but not sure if its correct



I want to be able to extend this out into say, if there was 3 items out of the 11 which were good, and the contestant had two goes to choose an item.






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 16:02









David G. Stork

11k41432




11k41432










asked Dec 13 '18 at 15:53









DH.DH.

44115




44115








  • 2




    $begingroup$
    That is far too low. If you just choose $1$ the probability that it is good is $frac 2{11}=18.18%$.
    $endgroup$
    – lulu
    Dec 13 '18 at 15:58






  • 1




    $begingroup$
    Your formulation is rather vague: "has the chance to win a prize if.."?... So picking the right one only gives a chance on winning a price? Secondly, what is a "turn" here?
    $endgroup$
    – drhab
    Dec 13 '18 at 15:59








  • 1




    $begingroup$
    Please write more clearly. What does "choose 2 from 3 out of 11 where 3 are good" mean?
    $endgroup$
    – lulu
    Dec 13 '18 at 15:59








  • 1




    $begingroup$
    If you are saying: there are three good ones and eight bad ones and you choose two, then the probability that both are bad is $frac 8{11}times frac 7{10}=frac {28}{55}approx 51%$. But is that what you meant?
    $endgroup$
    – lulu
    Dec 13 '18 at 16:01








  • 2




    $begingroup$
    Oh, you need both to be good? Ok, then it is $frac 3{11}times frac 2{10}=frac 3{55}$.
    $endgroup$
    – lulu
    Dec 13 '18 at 16:02














  • 2




    $begingroup$
    That is far too low. If you just choose $1$ the probability that it is good is $frac 2{11}=18.18%$.
    $endgroup$
    – lulu
    Dec 13 '18 at 15:58






  • 1




    $begingroup$
    Your formulation is rather vague: "has the chance to win a prize if.."?... So picking the right one only gives a chance on winning a price? Secondly, what is a "turn" here?
    $endgroup$
    – drhab
    Dec 13 '18 at 15:59








  • 1




    $begingroup$
    Please write more clearly. What does "choose 2 from 3 out of 11 where 3 are good" mean?
    $endgroup$
    – lulu
    Dec 13 '18 at 15:59








  • 1




    $begingroup$
    If you are saying: there are three good ones and eight bad ones and you choose two, then the probability that both are bad is $frac 8{11}times frac 7{10}=frac {28}{55}approx 51%$. But is that what you meant?
    $endgroup$
    – lulu
    Dec 13 '18 at 16:01








  • 2




    $begingroup$
    Oh, you need both to be good? Ok, then it is $frac 3{11}times frac 2{10}=frac 3{55}$.
    $endgroup$
    – lulu
    Dec 13 '18 at 16:02








2




2




$begingroup$
That is far too low. If you just choose $1$ the probability that it is good is $frac 2{11}=18.18%$.
$endgroup$
– lulu
Dec 13 '18 at 15:58




$begingroup$
That is far too low. If you just choose $1$ the probability that it is good is $frac 2{11}=18.18%$.
$endgroup$
– lulu
Dec 13 '18 at 15:58




1




1




$begingroup$
Your formulation is rather vague: "has the chance to win a prize if.."?... So picking the right one only gives a chance on winning a price? Secondly, what is a "turn" here?
$endgroup$
– drhab
Dec 13 '18 at 15:59






$begingroup$
Your formulation is rather vague: "has the chance to win a prize if.."?... So picking the right one only gives a chance on winning a price? Secondly, what is a "turn" here?
$endgroup$
– drhab
Dec 13 '18 at 15:59






1




1




$begingroup$
Please write more clearly. What does "choose 2 from 3 out of 11 where 3 are good" mean?
$endgroup$
– lulu
Dec 13 '18 at 15:59






$begingroup$
Please write more clearly. What does "choose 2 from 3 out of 11 where 3 are good" mean?
$endgroup$
– lulu
Dec 13 '18 at 15:59






1




1




$begingroup$
If you are saying: there are three good ones and eight bad ones and you choose two, then the probability that both are bad is $frac 8{11}times frac 7{10}=frac {28}{55}approx 51%$. But is that what you meant?
$endgroup$
– lulu
Dec 13 '18 at 16:01






$begingroup$
If you are saying: there are three good ones and eight bad ones and you choose two, then the probability that both are bad is $frac 8{11}times frac 7{10}=frac {28}{55}approx 51%$. But is that what you meant?
$endgroup$
– lulu
Dec 13 '18 at 16:01






2




2




$begingroup$
Oh, you need both to be good? Ok, then it is $frac 3{11}times frac 2{10}=frac 3{55}$.
$endgroup$
– lulu
Dec 13 '18 at 16:02




$begingroup$
Oh, you need both to be good? Ok, then it is $frac 3{11}times frac 2{10}=frac 3{55}$.
$endgroup$
– lulu
Dec 13 '18 at 16:02










2 Answers
2






active

oldest

votes


















1












$begingroup$

Say you have $n$ items out of which $m$ are good. In order to win, a person needs to draw a good item in the first turn, the probability of which is $frac mn$. In the next turn, the person again needs to draw a good item, which can happen with a probability of $frac{m-1}{n-1}$ since you have $m-1$ good items and $n-1$ total items left. The probability of winning by picking only good items in $m-1$ turns is thus the product,



$$displaystylefrac mncdotfrac{m-1}{n-1}cdotfrac{m-2}{n-2}ldotsfrac2{n-m+2}$$



You could have also found this probability by noting that you can select $m-1$ good items out of $m$ good ones in $binom m{m-1}=binom m1=m$ ways, and the total number of ways of drawing $m-1$ items out of $n$ items is $binom n{m-1}$. The required probability is $displaystylefrac m{binom n{m-1}}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If you only get one try, then



    $$P(win) = frac{good}{total} = frac{2}{11}$$



    If you get more than one try, its easier to calculate the compliment of not winning:



    $$P(win) = 1-P(loss)= 1-frac{9cdot 8 dots }{11cdot 10 dots}$$ where the right summand is the product of always losing for however many pulls you want.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Say you have $n$ items out of which $m$ are good. In order to win, a person needs to draw a good item in the first turn, the probability of which is $frac mn$. In the next turn, the person again needs to draw a good item, which can happen with a probability of $frac{m-1}{n-1}$ since you have $m-1$ good items and $n-1$ total items left. The probability of winning by picking only good items in $m-1$ turns is thus the product,



      $$displaystylefrac mncdotfrac{m-1}{n-1}cdotfrac{m-2}{n-2}ldotsfrac2{n-m+2}$$



      You could have also found this probability by noting that you can select $m-1$ good items out of $m$ good ones in $binom m{m-1}=binom m1=m$ ways, and the total number of ways of drawing $m-1$ items out of $n$ items is $binom n{m-1}$. The required probability is $displaystylefrac m{binom n{m-1}}$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Say you have $n$ items out of which $m$ are good. In order to win, a person needs to draw a good item in the first turn, the probability of which is $frac mn$. In the next turn, the person again needs to draw a good item, which can happen with a probability of $frac{m-1}{n-1}$ since you have $m-1$ good items and $n-1$ total items left. The probability of winning by picking only good items in $m-1$ turns is thus the product,



        $$displaystylefrac mncdotfrac{m-1}{n-1}cdotfrac{m-2}{n-2}ldotsfrac2{n-m+2}$$



        You could have also found this probability by noting that you can select $m-1$ good items out of $m$ good ones in $binom m{m-1}=binom m1=m$ ways, and the total number of ways of drawing $m-1$ items out of $n$ items is $binom n{m-1}$. The required probability is $displaystylefrac m{binom n{m-1}}$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Say you have $n$ items out of which $m$ are good. In order to win, a person needs to draw a good item in the first turn, the probability of which is $frac mn$. In the next turn, the person again needs to draw a good item, which can happen with a probability of $frac{m-1}{n-1}$ since you have $m-1$ good items and $n-1$ total items left. The probability of winning by picking only good items in $m-1$ turns is thus the product,



          $$displaystylefrac mncdotfrac{m-1}{n-1}cdotfrac{m-2}{n-2}ldotsfrac2{n-m+2}$$



          You could have also found this probability by noting that you can select $m-1$ good items out of $m$ good ones in $binom m{m-1}=binom m1=m$ ways, and the total number of ways of drawing $m-1$ items out of $n$ items is $binom n{m-1}$. The required probability is $displaystylefrac m{binom n{m-1}}$.






          share|cite|improve this answer









          $endgroup$



          Say you have $n$ items out of which $m$ are good. In order to win, a person needs to draw a good item in the first turn, the probability of which is $frac mn$. In the next turn, the person again needs to draw a good item, which can happen with a probability of $frac{m-1}{n-1}$ since you have $m-1$ good items and $n-1$ total items left. The probability of winning by picking only good items in $m-1$ turns is thus the product,



          $$displaystylefrac mncdotfrac{m-1}{n-1}cdotfrac{m-2}{n-2}ldotsfrac2{n-m+2}$$



          You could have also found this probability by noting that you can select $m-1$ good items out of $m$ good ones in $binom m{m-1}=binom m1=m$ ways, and the total number of ways of drawing $m-1$ items out of $n$ items is $binom n{m-1}$. The required probability is $displaystylefrac m{binom n{m-1}}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 16:18









          Shubham JohriShubham Johri

          5,189718




          5,189718























              0












              $begingroup$

              If you only get one try, then



              $$P(win) = frac{good}{total} = frac{2}{11}$$



              If you get more than one try, its easier to calculate the compliment of not winning:



              $$P(win) = 1-P(loss)= 1-frac{9cdot 8 dots }{11cdot 10 dots}$$ where the right summand is the product of always losing for however many pulls you want.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If you only get one try, then



                $$P(win) = frac{good}{total} = frac{2}{11}$$



                If you get more than one try, its easier to calculate the compliment of not winning:



                $$P(win) = 1-P(loss)= 1-frac{9cdot 8 dots }{11cdot 10 dots}$$ where the right summand is the product of always losing for however many pulls you want.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If you only get one try, then



                  $$P(win) = frac{good}{total} = frac{2}{11}$$



                  If you get more than one try, its easier to calculate the compliment of not winning:



                  $$P(win) = 1-P(loss)= 1-frac{9cdot 8 dots }{11cdot 10 dots}$$ where the right summand is the product of always losing for however many pulls you want.






                  share|cite|improve this answer









                  $endgroup$



                  If you only get one try, then



                  $$P(win) = frac{good}{total} = frac{2}{11}$$



                  If you get more than one try, its easier to calculate the compliment of not winning:



                  $$P(win) = 1-P(loss)= 1-frac{9cdot 8 dots }{11cdot 10 dots}$$ where the right summand is the product of always losing for however many pulls you want.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 16:03









                  David DiazDavid Diaz

                  979420




                  979420






























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