Consider an urn that contains $4$ blue and $7$ red balls












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I'm in need of some help with this problem. Consider an urn that contains $4$ blue and $7$ red balls. First one ball is selected, and then a second ball is selected without replacement.



(a) What is the probability that at least $1$ red ball is chosen?



$displaystylefrac{binom72cdotbinom41}{binom{11}2}=frac{28}{55}$



(b) What is the probability that the second ball chosen is red, given the that the first ball chosen was red?



$displaystylefrac{binom72}{binom{11}2}=frac{21}{55}$



(c) What is the probability that the second ball chosen is red, given that the first ball chosen was not red?



$displaystyle1-frac{binom71cdotbinom41}{binom{11}2}=frac{27}{55}$






probability probability-theory probability-distributions combinations







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  • 1




    $begingroup$
    I don't understand your computation for $(a)$. Easiest, I think, is to compute the complementary probability...the probability that both are blue.
    $endgroup$
    – lulu
    Dec 15 '18 at 21:43






  • 1




    $begingroup$
    For $(b)$...having chosen a red one first, there are now $6$ red ones among $10$ balls so the answer is...
    $endgroup$
    – lulu
    Dec 15 '18 at 21:45
















0












$begingroup$


I'm in need of some help with this problem. Consider an urn that contains $4$ blue and $7$ red balls. First one ball is selected, and then a second ball is selected without replacement.



(a) What is the probability that at least $1$ red ball is chosen?



$displaystylefrac{binom72cdotbinom41}{binom{11}2}=frac{28}{55}$



(b) What is the probability that the second ball chosen is red, given the that the first ball chosen was red?



$displaystylefrac{binom72}{binom{11}2}=frac{21}{55}$



(c) What is the probability that the second ball chosen is red, given that the first ball chosen was not red?



$displaystyle1-frac{binom71cdotbinom41}{binom{11}2}=frac{27}{55}$






probability probability-theory probability-distributions combinations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don't understand your computation for $(a)$. Easiest, I think, is to compute the complementary probability...the probability that both are blue.
    $endgroup$
    – lulu
    Dec 15 '18 at 21:43






  • 1




    $begingroup$
    For $(b)$...having chosen a red one first, there are now $6$ red ones among $10$ balls so the answer is...
    $endgroup$
    – lulu
    Dec 15 '18 at 21:45














0












0








0





$begingroup$


I'm in need of some help with this problem. Consider an urn that contains $4$ blue and $7$ red balls. First one ball is selected, and then a second ball is selected without replacement.



(a) What is the probability that at least $1$ red ball is chosen?



$displaystylefrac{binom72cdotbinom41}{binom{11}2}=frac{28}{55}$



(b) What is the probability that the second ball chosen is red, given the that the first ball chosen was red?



$displaystylefrac{binom72}{binom{11}2}=frac{21}{55}$



(c) What is the probability that the second ball chosen is red, given that the first ball chosen was not red?



$displaystyle1-frac{binom71cdotbinom41}{binom{11}2}=frac{27}{55}$






probability probability-theory probability-distributions combinations







share|cite|improve this question











$endgroup$




I'm in need of some help with this problem. Consider an urn that contains $4$ blue and $7$ red balls. First one ball is selected, and then a second ball is selected without replacement.



(a) What is the probability that at least $1$ red ball is chosen?



$displaystylefrac{binom72cdotbinom41}{binom{11}2}=frac{28}{55}$



(b) What is the probability that the second ball chosen is red, given the that the first ball chosen was red?



$displaystylefrac{binom72}{binom{11}2}=frac{21}{55}$



(c) What is the probability that the second ball chosen is red, given that the first ball chosen was not red?



$displaystyle1-frac{binom71cdotbinom41}{binom{11}2}=frac{27}{55}$






probability probability-theory probability-distributions combinations




probability probability-theory probability-distributions combinations






share|cite|improve this question















share|cite|improve this question













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edited Dec 16 '18 at 0:38









Key Flex

8,33261233




8,33261233










asked Dec 15 '18 at 21:41









Cole PhiperCole Phiper

11




11








  • 1




    $begingroup$
    I don't understand your computation for $(a)$. Easiest, I think, is to compute the complementary probability...the probability that both are blue.
    $endgroup$
    – lulu
    Dec 15 '18 at 21:43






  • 1




    $begingroup$
    For $(b)$...having chosen a red one first, there are now $6$ red ones among $10$ balls so the answer is...
    $endgroup$
    – lulu
    Dec 15 '18 at 21:45














  • 1




    $begingroup$
    I don't understand your computation for $(a)$. Easiest, I think, is to compute the complementary probability...the probability that both are blue.
    $endgroup$
    – lulu
    Dec 15 '18 at 21:43






  • 1




    $begingroup$
    For $(b)$...having chosen a red one first, there are now $6$ red ones among $10$ balls so the answer is...
    $endgroup$
    – lulu
    Dec 15 '18 at 21:45








1




1




$begingroup$
I don't understand your computation for $(a)$. Easiest, I think, is to compute the complementary probability...the probability that both are blue.
$endgroup$
– lulu
Dec 15 '18 at 21:43




$begingroup$
I don't understand your computation for $(a)$. Easiest, I think, is to compute the complementary probability...the probability that both are blue.
$endgroup$
– lulu
Dec 15 '18 at 21:43




1




1




$begingroup$
For $(b)$...having chosen a red one first, there are now $6$ red ones among $10$ balls so the answer is...
$endgroup$
– lulu
Dec 15 '18 at 21:45




$begingroup$
For $(b)$...having chosen a red one first, there are now $6$ red ones among $10$ balls so the answer is...
$endgroup$
– lulu
Dec 15 '18 at 21:45










2 Answers
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$begingroup$

(a) The probability that no red balls are chosen, i.e. the probability that two blue balls are chosen, is
$$frac{4}{11}frac{3}{10}$$
The probability that at least one red ball is chosen is therefore
$$1-frac{4}{11}frac{3}{10}$$



(b) If the first ball was chosen red, there are then $6$ more red balls out of $10$ total, so the probability of selecting a red ball second is $frac{6}{10}$



(c) Similar to (b). If the first ball chosen was blue, there are then $7$ red balls remaining and $10$ total, so the probability of selecting a red ball second is $frac{7}{10}$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    for the first part (A) you may ask: 2 balls are drawn from the urn (7 red, 4 blue) what is the probability that at least one is red (or in other words RR, OR ( RB/ BR )).



    so you can say, either: RR



    $displaystylefrac{binom72cdotbinom40}{binom{11}2}$



    OR (+) RB/BR



    $displaystylefrac{binom71cdotbinom41}{binom{11}2}$



    and together:



    $displaystylefrac{binom72cdotbinom40+binom71cdotbinom41}{binom{11}2}$



    for part b: you're left with 10 balls and you have 6 red - choose one.



    now you can paraphrase C based on B.



    Tip: this question falls under the hypergeometric probability category






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      (a) The probability that no red balls are chosen, i.e. the probability that two blue balls are chosen, is
      $$frac{4}{11}frac{3}{10}$$
      The probability that at least one red ball is chosen is therefore
      $$1-frac{4}{11}frac{3}{10}$$



      (b) If the first ball was chosen red, there are then $6$ more red balls out of $10$ total, so the probability of selecting a red ball second is $frac{6}{10}$



      (c) Similar to (b). If the first ball chosen was blue, there are then $7$ red balls remaining and $10$ total, so the probability of selecting a red ball second is $frac{7}{10}$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        (a) The probability that no red balls are chosen, i.e. the probability that two blue balls are chosen, is
        $$frac{4}{11}frac{3}{10}$$
        The probability that at least one red ball is chosen is therefore
        $$1-frac{4}{11}frac{3}{10}$$



        (b) If the first ball was chosen red, there are then $6$ more red balls out of $10$ total, so the probability of selecting a red ball second is $frac{6}{10}$



        (c) Similar to (b). If the first ball chosen was blue, there are then $7$ red balls remaining and $10$ total, so the probability of selecting a red ball second is $frac{7}{10}$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          (a) The probability that no red balls are chosen, i.e. the probability that two blue balls are chosen, is
          $$frac{4}{11}frac{3}{10}$$
          The probability that at least one red ball is chosen is therefore
          $$1-frac{4}{11}frac{3}{10}$$



          (b) If the first ball was chosen red, there are then $6$ more red balls out of $10$ total, so the probability of selecting a red ball second is $frac{6}{10}$



          (c) Similar to (b). If the first ball chosen was blue, there are then $7$ red balls remaining and $10$ total, so the probability of selecting a red ball second is $frac{7}{10}$






          share|cite|improve this answer









          $endgroup$



          (a) The probability that no red balls are chosen, i.e. the probability that two blue balls are chosen, is
          $$frac{4}{11}frac{3}{10}$$
          The probability that at least one red ball is chosen is therefore
          $$1-frac{4}{11}frac{3}{10}$$



          (b) If the first ball was chosen red, there are then $6$ more red balls out of $10$ total, so the probability of selecting a red ball second is $frac{6}{10}$



          (c) Similar to (b). If the first ball chosen was blue, there are then $7$ red balls remaining and $10$ total, so the probability of selecting a red ball second is $frac{7}{10}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 21:50









          pwerthpwerth

          3,265417




          3,265417























              1












              $begingroup$

              for the first part (A) you may ask: 2 balls are drawn from the urn (7 red, 4 blue) what is the probability that at least one is red (or in other words RR, OR ( RB/ BR )).



              so you can say, either: RR



              $displaystylefrac{binom72cdotbinom40}{binom{11}2}$



              OR (+) RB/BR



              $displaystylefrac{binom71cdotbinom41}{binom{11}2}$



              and together:



              $displaystylefrac{binom72cdotbinom40+binom71cdotbinom41}{binom{11}2}$



              for part b: you're left with 10 balls and you have 6 red - choose one.



              now you can paraphrase C based on B.



              Tip: this question falls under the hypergeometric probability category






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                for the first part (A) you may ask: 2 balls are drawn from the urn (7 red, 4 blue) what is the probability that at least one is red (or in other words RR, OR ( RB/ BR )).



                so you can say, either: RR



                $displaystylefrac{binom72cdotbinom40}{binom{11}2}$



                OR (+) RB/BR



                $displaystylefrac{binom71cdotbinom41}{binom{11}2}$



                and together:



                $displaystylefrac{binom72cdotbinom40+binom71cdotbinom41}{binom{11}2}$



                for part b: you're left with 10 balls and you have 6 red - choose one.



                now you can paraphrase C based on B.



                Tip: this question falls under the hypergeometric probability category






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  for the first part (A) you may ask: 2 balls are drawn from the urn (7 red, 4 blue) what is the probability that at least one is red (or in other words RR, OR ( RB/ BR )).



                  so you can say, either: RR



                  $displaystylefrac{binom72cdotbinom40}{binom{11}2}$



                  OR (+) RB/BR



                  $displaystylefrac{binom71cdotbinom41}{binom{11}2}$



                  and together:



                  $displaystylefrac{binom72cdotbinom40+binom71cdotbinom41}{binom{11}2}$



                  for part b: you're left with 10 balls and you have 6 red - choose one.



                  now you can paraphrase C based on B.



                  Tip: this question falls under the hypergeometric probability category






                  share|cite|improve this answer









                  $endgroup$



                  for the first part (A) you may ask: 2 balls are drawn from the urn (7 red, 4 blue) what is the probability that at least one is red (or in other words RR, OR ( RB/ BR )).



                  so you can say, either: RR



                  $displaystylefrac{binom72cdotbinom40}{binom{11}2}$



                  OR (+) RB/BR



                  $displaystylefrac{binom71cdotbinom41}{binom{11}2}$



                  and together:



                  $displaystylefrac{binom72cdotbinom40+binom71cdotbinom41}{binom{11}2}$



                  for part b: you're left with 10 balls and you have 6 red - choose one.



                  now you can paraphrase C based on B.



                  Tip: this question falls under the hypergeometric probability category







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 22:01









                  adhgadhg

                  340413




                  340413






























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