Drawing without replacement: why is the order of draw irrelevant? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)A question regarding drawing balls of differing colors from an urn before a certain number of draws occur without replacement.3 balls drawn from 1 urn - probability all same color (with/without replacement)Probability without replacement questionsBalls with and without replacementPicking balls blindfolded without replacementProbability of drawing balls without replacement in first and last drawAre expectation of with replacement and without replacement same? When?Choosing one type of ball without replacement.Drawing 4 balls from an urn without replacement and a bonus ballDrawing Balls Without Replacement

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Drawing without replacement: why is the order of draw irrelevant?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)A question regarding drawing balls of differing colors from an urn before a certain number of draws occur without replacement.3 balls drawn from 1 urn - probability all same color (with/without replacement)Probability without replacement questionsBalls with and without replacementPicking balls blindfolded without replacementProbability of drawing balls without replacement in first and last drawAre expectation of with replacement and without replacement same? When?Choosing one type of ball without replacement.Drawing 4 balls from an urn without replacement and a bonus ballDrawing Balls Without Replacement










3












$begingroup$


I am trying to wrap my head around this problem:




Daniel randomly chooses balls from the group of $6$ red and $4$ green. What is the probability that he picks $2$ red and $3$ green if balls are drawn without replacement.




What I remember from my college days that the probability is found by this formula:



$$P(A)=fracbinom62binom43binom105=frac521$$



Is this correct? I am trying to understand why this works. Wouldn't probability depend on the order of balls drawn as the number of balls is changing after each draw? I get how we obtain numerator and denominator, I just feel that the probability should be dependent on the order. For example, the probability to pick red first is $frac610$ so the probability for the second draw becomes $frac59$ for red and $frac49$ for green. But if the first picked ball is green, the probability for the second draw becomes $frac69$ for red and $frac39$ for green. What am I missing?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Usually the "nominator" is called numerator.
    $endgroup$
    – callculus
    10 hours ago










  • $begingroup$
    @callculus: yes, of, course, I need coffee :)
    $endgroup$
    – Vasya
    10 hours ago










  • $begingroup$
    The order doesn't matter because $P(A)$ is ultimately the sum of various conditional probabilities, and addition is both commutative and associative.
    $endgroup$
    – chepner
    8 hours ago















3












$begingroup$


I am trying to wrap my head around this problem:




Daniel randomly chooses balls from the group of $6$ red and $4$ green. What is the probability that he picks $2$ red and $3$ green if balls are drawn without replacement.




What I remember from my college days that the probability is found by this formula:



$$P(A)=fracbinom62binom43binom105=frac521$$



Is this correct? I am trying to understand why this works. Wouldn't probability depend on the order of balls drawn as the number of balls is changing after each draw? I get how we obtain numerator and denominator, I just feel that the probability should be dependent on the order. For example, the probability to pick red first is $frac610$ so the probability for the second draw becomes $frac59$ for red and $frac49$ for green. But if the first picked ball is green, the probability for the second draw becomes $frac69$ for red and $frac39$ for green. What am I missing?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Usually the "nominator" is called numerator.
    $endgroup$
    – callculus
    10 hours ago










  • $begingroup$
    @callculus: yes, of, course, I need coffee :)
    $endgroup$
    – Vasya
    10 hours ago










  • $begingroup$
    The order doesn't matter because $P(A)$ is ultimately the sum of various conditional probabilities, and addition is both commutative and associative.
    $endgroup$
    – chepner
    8 hours ago













3












3








3


3



$begingroup$


I am trying to wrap my head around this problem:




Daniel randomly chooses balls from the group of $6$ red and $4$ green. What is the probability that he picks $2$ red and $3$ green if balls are drawn without replacement.




What I remember from my college days that the probability is found by this formula:



$$P(A)=fracbinom62binom43binom105=frac521$$



Is this correct? I am trying to understand why this works. Wouldn't probability depend on the order of balls drawn as the number of balls is changing after each draw? I get how we obtain numerator and denominator, I just feel that the probability should be dependent on the order. For example, the probability to pick red first is $frac610$ so the probability for the second draw becomes $frac59$ for red and $frac49$ for green. But if the first picked ball is green, the probability for the second draw becomes $frac69$ for red and $frac39$ for green. What am I missing?










share|cite|improve this question











$endgroup$




I am trying to wrap my head around this problem:




Daniel randomly chooses balls from the group of $6$ red and $4$ green. What is the probability that he picks $2$ red and $3$ green if balls are drawn without replacement.




What I remember from my college days that the probability is found by this formula:



$$P(A)=fracbinom62binom43binom105=frac521$$



Is this correct? I am trying to understand why this works. Wouldn't probability depend on the order of balls drawn as the number of balls is changing after each draw? I get how we obtain numerator and denominator, I just feel that the probability should be dependent on the order. For example, the probability to pick red first is $frac610$ so the probability for the second draw becomes $frac59$ for red and $frac49$ for green. But if the first picked ball is green, the probability for the second draw becomes $frac69$ for red and $frac39$ for green. What am I missing?







probability probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 36 mins ago









JeffC

1053




1053










asked 10 hours ago









VasyaVasya

4,5441619




4,5441619







  • 1




    $begingroup$
    Usually the "nominator" is called numerator.
    $endgroup$
    – callculus
    10 hours ago










  • $begingroup$
    @callculus: yes, of, course, I need coffee :)
    $endgroup$
    – Vasya
    10 hours ago










  • $begingroup$
    The order doesn't matter because $P(A)$ is ultimately the sum of various conditional probabilities, and addition is both commutative and associative.
    $endgroup$
    – chepner
    8 hours ago












  • 1




    $begingroup$
    Usually the "nominator" is called numerator.
    $endgroup$
    – callculus
    10 hours ago










  • $begingroup$
    @callculus: yes, of, course, I need coffee :)
    $endgroup$
    – Vasya
    10 hours ago










  • $begingroup$
    The order doesn't matter because $P(A)$ is ultimately the sum of various conditional probabilities, and addition is both commutative and associative.
    $endgroup$
    – chepner
    8 hours ago







1




1




$begingroup$
Usually the "nominator" is called numerator.
$endgroup$
– callculus
10 hours ago




$begingroup$
Usually the "nominator" is called numerator.
$endgroup$
– callculus
10 hours ago












$begingroup$
@callculus: yes, of, course, I need coffee :)
$endgroup$
– Vasya
10 hours ago




$begingroup$
@callculus: yes, of, course, I need coffee :)
$endgroup$
– Vasya
10 hours ago












$begingroup$
The order doesn't matter because $P(A)$ is ultimately the sum of various conditional probabilities, and addition is both commutative and associative.
$endgroup$
– chepner
8 hours ago




$begingroup$
The order doesn't matter because $P(A)$ is ultimately the sum of various conditional probabilities, and addition is both commutative and associative.
$endgroup$
– chepner
8 hours ago










4 Answers
4






active

oldest

votes


















6












$begingroup$

If you took into consideration the order, you would get the same result but the calculation would be a bit more difficult:



  • all possible sequences of $5$ balls respecting order (as if they were distinguishable): $10cdot 9cdot 8cdot 7 cdot 6$

  • all possible selections of $colorred2$ out of $colorred6$ red balls: $colorredbinom62$

  • all possible selections of $colorgreen3$ out of $colorgreen4$ green balls: $colorgreenbinom43$

  • all possible arrangements of the selected $colorred2+colorgreen3$ balls: $5!$

All together
$$fraccolorredbinom62cdot colorgreenbinom43 cdot 5!10cdot 9cdot 8cdot 7 cdot 6 = fraccolorredbinom62cdot colorgreenbinom43frac10!5!cdot 5!= frac521$$






share|cite|improve this answer









$endgroup$




















    6












    $begingroup$

    The probability of picking a red ball first and then a green ball is
    $$ frac610 cdot frac49 $$
    The probability of picking a green ball first and then a red ball is
    $$ frac410 cdot frac69 $$
    Notice that the numbers in the denominator are the same, while the numbers in
    the numerator are the same but in reverse order? Multiplication is commutative.



    Another way of looking at this: we don't care about the process you go through in picking the balls, as long as it is fair: each possible outcome (i.e. each possible subset of
    $5$ of the $10$ balls, where we consider the balls as in principle distinguishable) has the same probability. If this is the case, you just need to count the number of
    outcomes that belong to the event you're considering, and divide by the total number of
    outcomes.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective!
      $endgroup$
      – Vasya
      10 hours ago


















    2












    $begingroup$

    You can comprehend the calculation in a simpler way with smaller numbers.




    Daniel randomly chooses balls from the group of $3$ red and $2$ green. What
    is the probability that he picks $2$ red and $2$ green if balls are drawn
    without replacement.




    Indeed we have to regard the order. There are $frac4!2!cdot 2!=6$ ways to draw 2 red and 2 green balls:



    $$colorgreengcolorgreengcolorredrcolorredr, colorgreengcolorredrcolorgreengcolorredr, colorgreengcolorredrcolorredrcolorgreeng, colorredrcolorgreengcolorgreengcolorredr, colorredrcolorgreengcolorredrcolorgreeng, colorredrcolorredrcolorgreengcolorgreeng$$



    Each way has the same probability: $frac35cdot frac24cdot frac23cdot frac12 quad (ggrr)$



    Multiplying with 6 (ways) we get $6cdot frac35cdot frac24cdot frac23cdot frac12=frac35=0.6 $



    Using binomial coefficients we get $fracbinom32cdot binom22binom54=frac3cdot 15=frac35=0.6$



    And we get the same result.






    share|cite|improve this answer











    $endgroup$




















      1












      $begingroup$

      There is a principle called "conservation of expected evidence" that says that if you have events A and B, then when you calculate the probability of A without knowing whether B happens, the result should be the same as the expected value of the probability over the possible results of B.



      In this case, let A be the probability that the second ball is red, and B be the probability the first one is green. The principle says that P(A) = P(A|B)P(B)+P(A|~B)P(~B). That is, if you split A into two cases of A and B versus A and not B, the total probability should just be the probability of A. If you roll a die and flip a coin, the probability of getting a 1 one the die should change if you split it into P(die=1,coin=heads) plus P(die=1,coin=tails).



      We have the following values for those probabilities:



      P(A) = $frac610$

      P(A|B) = $frac5 9 $

      P(B) = $frac610$

      P(A|~B) = $frac 6 9 $

      P(~B) = $frac 4 10$



      So the equation is $frac610 = frac5 9 frac610+frac 6 9 frac 4 10=frac30+249*10 = frac549*10=frac9*69*10=frac 6 10$



      If you have ten cards, 6 red and 4 green, and you shuffle them, would the probability of the first one being red be any different from the probability of the second one being red?






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        If you took into consideration the order, you would get the same result but the calculation would be a bit more difficult:



        • all possible sequences of $5$ balls respecting order (as if they were distinguishable): $10cdot 9cdot 8cdot 7 cdot 6$

        • all possible selections of $colorred2$ out of $colorred6$ red balls: $colorredbinom62$

        • all possible selections of $colorgreen3$ out of $colorgreen4$ green balls: $colorgreenbinom43$

        • all possible arrangements of the selected $colorred2+colorgreen3$ balls: $5!$

        All together
        $$fraccolorredbinom62cdot colorgreenbinom43 cdot 5!10cdot 9cdot 8cdot 7 cdot 6 = fraccolorredbinom62cdot colorgreenbinom43frac10!5!cdot 5!= frac521$$






        share|cite|improve this answer









        $endgroup$

















          6












          $begingroup$

          If you took into consideration the order, you would get the same result but the calculation would be a bit more difficult:



          • all possible sequences of $5$ balls respecting order (as if they were distinguishable): $10cdot 9cdot 8cdot 7 cdot 6$

          • all possible selections of $colorred2$ out of $colorred6$ red balls: $colorredbinom62$

          • all possible selections of $colorgreen3$ out of $colorgreen4$ green balls: $colorgreenbinom43$

          • all possible arrangements of the selected $colorred2+colorgreen3$ balls: $5!$

          All together
          $$fraccolorredbinom62cdot colorgreenbinom43 cdot 5!10cdot 9cdot 8cdot 7 cdot 6 = fraccolorredbinom62cdot colorgreenbinom43frac10!5!cdot 5!= frac521$$






          share|cite|improve this answer









          $endgroup$















            6












            6








            6





            $begingroup$

            If you took into consideration the order, you would get the same result but the calculation would be a bit more difficult:



            • all possible sequences of $5$ balls respecting order (as if they were distinguishable): $10cdot 9cdot 8cdot 7 cdot 6$

            • all possible selections of $colorred2$ out of $colorred6$ red balls: $colorredbinom62$

            • all possible selections of $colorgreen3$ out of $colorgreen4$ green balls: $colorgreenbinom43$

            • all possible arrangements of the selected $colorred2+colorgreen3$ balls: $5!$

            All together
            $$fraccolorredbinom62cdot colorgreenbinom43 cdot 5!10cdot 9cdot 8cdot 7 cdot 6 = fraccolorredbinom62cdot colorgreenbinom43frac10!5!cdot 5!= frac521$$






            share|cite|improve this answer









            $endgroup$



            If you took into consideration the order, you would get the same result but the calculation would be a bit more difficult:



            • all possible sequences of $5$ balls respecting order (as if they were distinguishable): $10cdot 9cdot 8cdot 7 cdot 6$

            • all possible selections of $colorred2$ out of $colorred6$ red balls: $colorredbinom62$

            • all possible selections of $colorgreen3$ out of $colorgreen4$ green balls: $colorgreenbinom43$

            • all possible arrangements of the selected $colorred2+colorgreen3$ balls: $5!$

            All together
            $$fraccolorredbinom62cdot colorgreenbinom43 cdot 5!10cdot 9cdot 8cdot 7 cdot 6 = fraccolorredbinom62cdot colorgreenbinom43frac10!5!cdot 5!= frac521$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 10 hours ago









            trancelocationtrancelocation

            14.5k1929




            14.5k1929





















                6












                $begingroup$

                The probability of picking a red ball first and then a green ball is
                $$ frac610 cdot frac49 $$
                The probability of picking a green ball first and then a red ball is
                $$ frac410 cdot frac69 $$
                Notice that the numbers in the denominator are the same, while the numbers in
                the numerator are the same but in reverse order? Multiplication is commutative.



                Another way of looking at this: we don't care about the process you go through in picking the balls, as long as it is fair: each possible outcome (i.e. each possible subset of
                $5$ of the $10$ balls, where we consider the balls as in principle distinguishable) has the same probability. If this is the case, you just need to count the number of
                outcomes that belong to the event you're considering, and divide by the total number of
                outcomes.






                share|cite|improve this answer









                $endgroup$












                • $begingroup$
                  Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective!
                  $endgroup$
                  – Vasya
                  10 hours ago















                6












                $begingroup$

                The probability of picking a red ball first and then a green ball is
                $$ frac610 cdot frac49 $$
                The probability of picking a green ball first and then a red ball is
                $$ frac410 cdot frac69 $$
                Notice that the numbers in the denominator are the same, while the numbers in
                the numerator are the same but in reverse order? Multiplication is commutative.



                Another way of looking at this: we don't care about the process you go through in picking the balls, as long as it is fair: each possible outcome (i.e. each possible subset of
                $5$ of the $10$ balls, where we consider the balls as in principle distinguishable) has the same probability. If this is the case, you just need to count the number of
                outcomes that belong to the event you're considering, and divide by the total number of
                outcomes.






                share|cite|improve this answer









                $endgroup$












                • $begingroup$
                  Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective!
                  $endgroup$
                  – Vasya
                  10 hours ago













                6












                6








                6





                $begingroup$

                The probability of picking a red ball first and then a green ball is
                $$ frac610 cdot frac49 $$
                The probability of picking a green ball first and then a red ball is
                $$ frac410 cdot frac69 $$
                Notice that the numbers in the denominator are the same, while the numbers in
                the numerator are the same but in reverse order? Multiplication is commutative.



                Another way of looking at this: we don't care about the process you go through in picking the balls, as long as it is fair: each possible outcome (i.e. each possible subset of
                $5$ of the $10$ balls, where we consider the balls as in principle distinguishable) has the same probability. If this is the case, you just need to count the number of
                outcomes that belong to the event you're considering, and divide by the total number of
                outcomes.






                share|cite|improve this answer









                $endgroup$



                The probability of picking a red ball first and then a green ball is
                $$ frac610 cdot frac49 $$
                The probability of picking a green ball first and then a red ball is
                $$ frac410 cdot frac69 $$
                Notice that the numbers in the denominator are the same, while the numbers in
                the numerator are the same but in reverse order? Multiplication is commutative.



                Another way of looking at this: we don't care about the process you go through in picking the balls, as long as it is fair: each possible outcome (i.e. each possible subset of
                $5$ of the $10$ balls, where we consider the balls as in principle distinguishable) has the same probability. If this is the case, you just need to count the number of
                outcomes that belong to the event you're considering, and divide by the total number of
                outcomes.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 10 hours ago









                Robert IsraelRobert Israel

                332k23222481




                332k23222481











                • $begingroup$
                  Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective!
                  $endgroup$
                  – Vasya
                  10 hours ago
















                • $begingroup$
                  Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective!
                  $endgroup$
                  – Vasya
                  10 hours ago















                $begingroup$
                Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective!
                $endgroup$
                – Vasya
                10 hours ago




                $begingroup$
                Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective!
                $endgroup$
                – Vasya
                10 hours ago











                2












                $begingroup$

                You can comprehend the calculation in a simpler way with smaller numbers.




                Daniel randomly chooses balls from the group of $3$ red and $2$ green. What
                is the probability that he picks $2$ red and $2$ green if balls are drawn
                without replacement.




                Indeed we have to regard the order. There are $frac4!2!cdot 2!=6$ ways to draw 2 red and 2 green balls:



                $$colorgreengcolorgreengcolorredrcolorredr, colorgreengcolorredrcolorgreengcolorredr, colorgreengcolorredrcolorredrcolorgreeng, colorredrcolorgreengcolorgreengcolorredr, colorredrcolorgreengcolorredrcolorgreeng, colorredrcolorredrcolorgreengcolorgreeng$$



                Each way has the same probability: $frac35cdot frac24cdot frac23cdot frac12 quad (ggrr)$



                Multiplying with 6 (ways) we get $6cdot frac35cdot frac24cdot frac23cdot frac12=frac35=0.6 $



                Using binomial coefficients we get $fracbinom32cdot binom22binom54=frac3cdot 15=frac35=0.6$



                And we get the same result.






                share|cite|improve this answer











                $endgroup$

















                  2












                  $begingroup$

                  You can comprehend the calculation in a simpler way with smaller numbers.




                  Daniel randomly chooses balls from the group of $3$ red and $2$ green. What
                  is the probability that he picks $2$ red and $2$ green if balls are drawn
                  without replacement.




                  Indeed we have to regard the order. There are $frac4!2!cdot 2!=6$ ways to draw 2 red and 2 green balls:



                  $$colorgreengcolorgreengcolorredrcolorredr, colorgreengcolorredrcolorgreengcolorredr, colorgreengcolorredrcolorredrcolorgreeng, colorredrcolorgreengcolorgreengcolorredr, colorredrcolorgreengcolorredrcolorgreeng, colorredrcolorredrcolorgreengcolorgreeng$$



                  Each way has the same probability: $frac35cdot frac24cdot frac23cdot frac12 quad (ggrr)$



                  Multiplying with 6 (ways) we get $6cdot frac35cdot frac24cdot frac23cdot frac12=frac35=0.6 $



                  Using binomial coefficients we get $fracbinom32cdot binom22binom54=frac3cdot 15=frac35=0.6$



                  And we get the same result.






                  share|cite|improve this answer











                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    You can comprehend the calculation in a simpler way with smaller numbers.




                    Daniel randomly chooses balls from the group of $3$ red and $2$ green. What
                    is the probability that he picks $2$ red and $2$ green if balls are drawn
                    without replacement.




                    Indeed we have to regard the order. There are $frac4!2!cdot 2!=6$ ways to draw 2 red and 2 green balls:



                    $$colorgreengcolorgreengcolorredrcolorredr, colorgreengcolorredrcolorgreengcolorredr, colorgreengcolorredrcolorredrcolorgreeng, colorredrcolorgreengcolorgreengcolorredr, colorredrcolorgreengcolorredrcolorgreeng, colorredrcolorredrcolorgreengcolorgreeng$$



                    Each way has the same probability: $frac35cdot frac24cdot frac23cdot frac12 quad (ggrr)$



                    Multiplying with 6 (ways) we get $6cdot frac35cdot frac24cdot frac23cdot frac12=frac35=0.6 $



                    Using binomial coefficients we get $fracbinom32cdot binom22binom54=frac3cdot 15=frac35=0.6$



                    And we get the same result.






                    share|cite|improve this answer











                    $endgroup$



                    You can comprehend the calculation in a simpler way with smaller numbers.




                    Daniel randomly chooses balls from the group of $3$ red and $2$ green. What
                    is the probability that he picks $2$ red and $2$ green if balls are drawn
                    without replacement.




                    Indeed we have to regard the order. There are $frac4!2!cdot 2!=6$ ways to draw 2 red and 2 green balls:



                    $$colorgreengcolorgreengcolorredrcolorredr, colorgreengcolorredrcolorgreengcolorredr, colorgreengcolorredrcolorredrcolorgreeng, colorredrcolorgreengcolorgreengcolorredr, colorredrcolorgreengcolorredrcolorgreeng, colorredrcolorredrcolorgreengcolorgreeng$$



                    Each way has the same probability: $frac35cdot frac24cdot frac23cdot frac12 quad (ggrr)$



                    Multiplying with 6 (ways) we get $6cdot frac35cdot frac24cdot frac23cdot frac12=frac35=0.6 $



                    Using binomial coefficients we get $fracbinom32cdot binom22binom54=frac3cdot 15=frac35=0.6$



                    And we get the same result.







                    share|cite|improve this answer














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                    edited 10 hours ago

























                    answered 10 hours ago









                    callculuscallculus

                    18.8k31528




                    18.8k31528





















                        1












                        $begingroup$

                        There is a principle called "conservation of expected evidence" that says that if you have events A and B, then when you calculate the probability of A without knowing whether B happens, the result should be the same as the expected value of the probability over the possible results of B.



                        In this case, let A be the probability that the second ball is red, and B be the probability the first one is green. The principle says that P(A) = P(A|B)P(B)+P(A|~B)P(~B). That is, if you split A into two cases of A and B versus A and not B, the total probability should just be the probability of A. If you roll a die and flip a coin, the probability of getting a 1 one the die should change if you split it into P(die=1,coin=heads) plus P(die=1,coin=tails).



                        We have the following values for those probabilities:



                        P(A) = $frac610$

                        P(A|B) = $frac5 9 $

                        P(B) = $frac610$

                        P(A|~B) = $frac 6 9 $

                        P(~B) = $frac 4 10$



                        So the equation is $frac610 = frac5 9 frac610+frac 6 9 frac 4 10=frac30+249*10 = frac549*10=frac9*69*10=frac 6 10$



                        If you have ten cards, 6 red and 4 green, and you shuffle them, would the probability of the first one being red be any different from the probability of the second one being red?






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          There is a principle called "conservation of expected evidence" that says that if you have events A and B, then when you calculate the probability of A without knowing whether B happens, the result should be the same as the expected value of the probability over the possible results of B.



                          In this case, let A be the probability that the second ball is red, and B be the probability the first one is green. The principle says that P(A) = P(A|B)P(B)+P(A|~B)P(~B). That is, if you split A into two cases of A and B versus A and not B, the total probability should just be the probability of A. If you roll a die and flip a coin, the probability of getting a 1 one the die should change if you split it into P(die=1,coin=heads) plus P(die=1,coin=tails).



                          We have the following values for those probabilities:



                          P(A) = $frac610$

                          P(A|B) = $frac5 9 $

                          P(B) = $frac610$

                          P(A|~B) = $frac 6 9 $

                          P(~B) = $frac 4 10$



                          So the equation is $frac610 = frac5 9 frac610+frac 6 9 frac 4 10=frac30+249*10 = frac549*10=frac9*69*10=frac 6 10$



                          If you have ten cards, 6 red and 4 green, and you shuffle them, would the probability of the first one being red be any different from the probability of the second one being red?






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            There is a principle called "conservation of expected evidence" that says that if you have events A and B, then when you calculate the probability of A without knowing whether B happens, the result should be the same as the expected value of the probability over the possible results of B.



                            In this case, let A be the probability that the second ball is red, and B be the probability the first one is green. The principle says that P(A) = P(A|B)P(B)+P(A|~B)P(~B). That is, if you split A into two cases of A and B versus A and not B, the total probability should just be the probability of A. If you roll a die and flip a coin, the probability of getting a 1 one the die should change if you split it into P(die=1,coin=heads) plus P(die=1,coin=tails).



                            We have the following values for those probabilities:



                            P(A) = $frac610$

                            P(A|B) = $frac5 9 $

                            P(B) = $frac610$

                            P(A|~B) = $frac 6 9 $

                            P(~B) = $frac 4 10$



                            So the equation is $frac610 = frac5 9 frac610+frac 6 9 frac 4 10=frac30+249*10 = frac549*10=frac9*69*10=frac 6 10$



                            If you have ten cards, 6 red and 4 green, and you shuffle them, would the probability of the first one being red be any different from the probability of the second one being red?






                            share|cite|improve this answer









                            $endgroup$



                            There is a principle called "conservation of expected evidence" that says that if you have events A and B, then when you calculate the probability of A without knowing whether B happens, the result should be the same as the expected value of the probability over the possible results of B.



                            In this case, let A be the probability that the second ball is red, and B be the probability the first one is green. The principle says that P(A) = P(A|B)P(B)+P(A|~B)P(~B). That is, if you split A into two cases of A and B versus A and not B, the total probability should just be the probability of A. If you roll a die and flip a coin, the probability of getting a 1 one the die should change if you split it into P(die=1,coin=heads) plus P(die=1,coin=tails).



                            We have the following values for those probabilities:



                            P(A) = $frac610$

                            P(A|B) = $frac5 9 $

                            P(B) = $frac610$

                            P(A|~B) = $frac 6 9 $

                            P(~B) = $frac 4 10$



                            So the equation is $frac610 = frac5 9 frac610+frac 6 9 frac 4 10=frac30+249*10 = frac549*10=frac9*69*10=frac 6 10$



                            If you have ten cards, 6 red and 4 green, and you shuffle them, would the probability of the first one being red be any different from the probability of the second one being red?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 7 hours ago









                            AcccumulationAcccumulation

                            7,3232619




                            7,3232619



























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