A box has $4$ red and $20$ white balls. A person takes $10$ balls. What is the probability that all or none of the red balls were taken?Box A has 3 red and 7 white ballsWe have two boxes, in box #1 there are 7 white balls and 5 red balls…probabilityA box contains 5 blue balls, 4 red balls and 9 green balls. A person takes out 3 balls from the box.An urn has 3 red balls 4 white balls and 2 black ballsConditional probability on red and white ballsBoth balls are red. What is the probability that they both came from the same box?Suppose there are 9 white balls and 3 red balls. A,B,C take turns to pick a ball without replacement . The first person to get a red ball wins.What is the probability that the person picks up all three red balls?drawing 2 red balls from box contains 3 red balls and 7 white ballsA box contains 3 red balls and 2 white balls

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A box has $4$ red and $20$ white balls. A person takes $10$ balls. What is the probability that all or none of the red balls were taken?


Box A has 3 red and 7 white ballsWe have two boxes, in box #1 there are 7 white balls and 5 red balls…probabilityA box contains 5 blue balls, 4 red balls and 9 green balls. A person takes out 3 balls from the box.An urn has 3 red balls 4 white balls and 2 black ballsConditional probability on red and white ballsBoth balls are red. What is the probability that they both came from the same box?Suppose there are 9 white balls and 3 red balls. A,B,C take turns to pick a ball without replacement . The first person to get a red ball wins.What is the probability that the person picks up all three red balls?drawing 2 red balls from box contains 3 red balls and 7 white ballsA box contains 3 red balls and 2 white balls













3












$begingroup$


A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red ones?



I don't understand why is this correct.



$$frac20 choose 6+20 choose 1024 choose 10 $$










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red ones?



    I don't understand why is this correct.



    $$frac20 choose 6+20 choose 1024 choose 10 $$










    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red ones?



      I don't understand why is this correct.



      $$frac20 choose 6+20 choose 1024 choose 10 $$










      share|cite|improve this question











      $endgroup$




      A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red ones?



      I don't understand why is this correct.



      $$frac20 choose 6+20 choose 1024 choose 10 $$







      probability combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      user

      5,84011031




      5,84011031










      asked 2 days ago









      GuidoGuido

      355




      355




















          4 Answers
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          active

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          4












          $begingroup$

          Because $tfracbinom 44binom206binom1414binom2410binom1414$ is the probability that the first person select 4 red and 6 white and the second person receives the remaining 14 white, while $tfracbinom2010binom44binom1010binom2410binom1414$ is the probability that the first person selects 10 white and second person select 4 red and 10 white.



          $$~\dfracbinom 44binom206binom1414+binom2010binom44binom1010binom2410binom1414=dfracbinom206+binom2010binom2410$$






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            I would assume that $binom204$ in the final expression is a typo...
            $endgroup$
            – user
            2 days ago










          • $begingroup$
            Yas. Good catch.
            $endgroup$
            – Graham Kemp
            2 days ago


















          4












          $begingroup$

          You can use the Hypergeometric distribution.



          a) The probability that the first person pick up 4 red ones is $fracbinom44cdot binom206binom2410$



          Then the (conditional) probability that the second person does not pick up 4 red ones is just $1.$



          b) The probability that the first person does not pick up 4 red ones is $fracbinom40cdot binom2010binom2410$



          The (conditional) probability that the second person pick up 4 red one is $fracbinom44cdot binom1010binom1414=1$



          So in total we have



          $$fracbinom44cdot binom206binom2410+fracbinom40cdot binom2010binom2410=frac binom206binom2410+frac binom2010binom2410=fracbinom206+ binom2010binom2410$$






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            There are two ways for a "good event" to occur in this probability space: either player one picks up all 4 red balls (first type of good event), or player 1 picks up none of the red balls (second type of good event; equivalent to saying player 2 picks up all red balls).



            The total number of ways that the first type of good event can occur is the number of ways we can pick 10 - 4 = 6 balls amond 24 - 4 = 20 balls after we fix that player one has selected the 4 red balls. In other words, there are $20 choose 6$ ways the first type of good event can occur.



            The total number of ways that the second type of good event may occur is the number of ways player one can pick all of his balls exclusively from the white balls. Since there are 20 white balls and player 1 picks 10 balls in total, this gives us $20 choose 10$ ways for the second type of good event to occur.



            Since the two "good" event spaces are mutually exclusive (player 1 can't pick all the red balls and none of the red balls at the same time), we can add them up obtain the total number of good event (the numerator in the solution you provided), and divide this by the size of the probability space (the total number of ways player 1 can pick 10 balls from 24, which is your denominator term).



            Does this answer your question?






            share|cite|improve this answer








            New contributor




            ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$




















              1












              $begingroup$

              Firstly, note that the events are mutually exclusive events. Therefore, the probability overall is the sum of the probability of the first person and second person selecting 4 red balls.



              The number of ways the first person selects the 4 red balls is the same as the number of ways they can select the remaining six as white balls (they have 10 chances in total) from the 20 white balls available. This can be done in $24-4choose6$ ways



              The number of ways the second person selects the 4 red balls is the same as the number of ways they can select the remaining ten as white balls (they have 14 chances on total) from the 20 white balls available. This can be done in $24-4choose10$ ways.



              The total number of ways to split the 24 balls is $24choose10$



              Therefore, the probability is what you put down.






              share|cite|improve this answer










              New contributor




              aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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              $endgroup$












                Your Answer





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                4 Answers
                4






                active

                oldest

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                4 Answers
                4






                active

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                active

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                $begingroup$

                Because $tfracbinom 44binom206binom1414binom2410binom1414$ is the probability that the first person select 4 red and 6 white and the second person receives the remaining 14 white, while $tfracbinom2010binom44binom1010binom2410binom1414$ is the probability that the first person selects 10 white and second person select 4 red and 10 white.



                $$~\dfracbinom 44binom206binom1414+binom2010binom44binom1010binom2410binom1414=dfracbinom206+binom2010binom2410$$






                share|cite|improve this answer











                $endgroup$








                • 1




                  $begingroup$
                  I would assume that $binom204$ in the final expression is a typo...
                  $endgroup$
                  – user
                  2 days ago










                • $begingroup$
                  Yas. Good catch.
                  $endgroup$
                  – Graham Kemp
                  2 days ago















                4












                $begingroup$

                Because $tfracbinom 44binom206binom1414binom2410binom1414$ is the probability that the first person select 4 red and 6 white and the second person receives the remaining 14 white, while $tfracbinom2010binom44binom1010binom2410binom1414$ is the probability that the first person selects 10 white and second person select 4 red and 10 white.



                $$~\dfracbinom 44binom206binom1414+binom2010binom44binom1010binom2410binom1414=dfracbinom206+binom2010binom2410$$






                share|cite|improve this answer











                $endgroup$








                • 1




                  $begingroup$
                  I would assume that $binom204$ in the final expression is a typo...
                  $endgroup$
                  – user
                  2 days ago










                • $begingroup$
                  Yas. Good catch.
                  $endgroup$
                  – Graham Kemp
                  2 days ago













                4












                4








                4





                $begingroup$

                Because $tfracbinom 44binom206binom1414binom2410binom1414$ is the probability that the first person select 4 red and 6 white and the second person receives the remaining 14 white, while $tfracbinom2010binom44binom1010binom2410binom1414$ is the probability that the first person selects 10 white and second person select 4 red and 10 white.



                $$~\dfracbinom 44binom206binom1414+binom2010binom44binom1010binom2410binom1414=dfracbinom206+binom2010binom2410$$






                share|cite|improve this answer











                $endgroup$



                Because $tfracbinom 44binom206binom1414binom2410binom1414$ is the probability that the first person select 4 red and 6 white and the second person receives the remaining 14 white, while $tfracbinom2010binom44binom1010binom2410binom1414$ is the probability that the first person selects 10 white and second person select 4 red and 10 white.



                $$~\dfracbinom 44binom206binom1414+binom2010binom44binom1010binom2410binom1414=dfracbinom206+binom2010binom2410$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered 2 days ago









                Graham KempGraham Kemp

                87.3k43579




                87.3k43579







                • 1




                  $begingroup$
                  I would assume that $binom204$ in the final expression is a typo...
                  $endgroup$
                  – user
                  2 days ago










                • $begingroup$
                  Yas. Good catch.
                  $endgroup$
                  – Graham Kemp
                  2 days ago












                • 1




                  $begingroup$
                  I would assume that $binom204$ in the final expression is a typo...
                  $endgroup$
                  – user
                  2 days ago










                • $begingroup$
                  Yas. Good catch.
                  $endgroup$
                  – Graham Kemp
                  2 days ago







                1




                1




                $begingroup$
                I would assume that $binom204$ in the final expression is a typo...
                $endgroup$
                – user
                2 days ago




                $begingroup$
                I would assume that $binom204$ in the final expression is a typo...
                $endgroup$
                – user
                2 days ago












                $begingroup$
                Yas. Good catch.
                $endgroup$
                – Graham Kemp
                2 days ago




                $begingroup$
                Yas. Good catch.
                $endgroup$
                – Graham Kemp
                2 days ago











                4












                $begingroup$

                You can use the Hypergeometric distribution.



                a) The probability that the first person pick up 4 red ones is $fracbinom44cdot binom206binom2410$



                Then the (conditional) probability that the second person does not pick up 4 red ones is just $1.$



                b) The probability that the first person does not pick up 4 red ones is $fracbinom40cdot binom2010binom2410$



                The (conditional) probability that the second person pick up 4 red one is $fracbinom44cdot binom1010binom1414=1$



                So in total we have



                $$fracbinom44cdot binom206binom2410+fracbinom40cdot binom2010binom2410=frac binom206binom2410+frac binom2010binom2410=fracbinom206+ binom2010binom2410$$






                share|cite|improve this answer









                $endgroup$

















                  4












                  $begingroup$

                  You can use the Hypergeometric distribution.



                  a) The probability that the first person pick up 4 red ones is $fracbinom44cdot binom206binom2410$



                  Then the (conditional) probability that the second person does not pick up 4 red ones is just $1.$



                  b) The probability that the first person does not pick up 4 red ones is $fracbinom40cdot binom2010binom2410$



                  The (conditional) probability that the second person pick up 4 red one is $fracbinom44cdot binom1010binom1414=1$



                  So in total we have



                  $$fracbinom44cdot binom206binom2410+fracbinom40cdot binom2010binom2410=frac binom206binom2410+frac binom2010binom2410=fracbinom206+ binom2010binom2410$$






                  share|cite|improve this answer









                  $endgroup$















                    4












                    4








                    4





                    $begingroup$

                    You can use the Hypergeometric distribution.



                    a) The probability that the first person pick up 4 red ones is $fracbinom44cdot binom206binom2410$



                    Then the (conditional) probability that the second person does not pick up 4 red ones is just $1.$



                    b) The probability that the first person does not pick up 4 red ones is $fracbinom40cdot binom2010binom2410$



                    The (conditional) probability that the second person pick up 4 red one is $fracbinom44cdot binom1010binom1414=1$



                    So in total we have



                    $$fracbinom44cdot binom206binom2410+fracbinom40cdot binom2010binom2410=frac binom206binom2410+frac binom2010binom2410=fracbinom206+ binom2010binom2410$$






                    share|cite|improve this answer









                    $endgroup$



                    You can use the Hypergeometric distribution.



                    a) The probability that the first person pick up 4 red ones is $fracbinom44cdot binom206binom2410$



                    Then the (conditional) probability that the second person does not pick up 4 red ones is just $1.$



                    b) The probability that the first person does not pick up 4 red ones is $fracbinom40cdot binom2010binom2410$



                    The (conditional) probability that the second person pick up 4 red one is $fracbinom44cdot binom1010binom1414=1$



                    So in total we have



                    $$fracbinom44cdot binom206binom2410+fracbinom40cdot binom2010binom2410=frac binom206binom2410+frac binom2010binom2410=fracbinom206+ binom2010binom2410$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    callculuscallculus

                    18.6k31428




                    18.6k31428





















                        1












                        $begingroup$

                        There are two ways for a "good event" to occur in this probability space: either player one picks up all 4 red balls (first type of good event), or player 1 picks up none of the red balls (second type of good event; equivalent to saying player 2 picks up all red balls).



                        The total number of ways that the first type of good event can occur is the number of ways we can pick 10 - 4 = 6 balls amond 24 - 4 = 20 balls after we fix that player one has selected the 4 red balls. In other words, there are $20 choose 6$ ways the first type of good event can occur.



                        The total number of ways that the second type of good event may occur is the number of ways player one can pick all of his balls exclusively from the white balls. Since there are 20 white balls and player 1 picks 10 balls in total, this gives us $20 choose 10$ ways for the second type of good event to occur.



                        Since the two "good" event spaces are mutually exclusive (player 1 can't pick all the red balls and none of the red balls at the same time), we can add them up obtain the total number of good event (the numerator in the solution you provided), and divide this by the size of the probability space (the total number of ways player 1 can pick 10 balls from 24, which is your denominator term).



                        Does this answer your question?






                        share|cite|improve this answer








                        New contributor




                        ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






                        $endgroup$

















                          1












                          $begingroup$

                          There are two ways for a "good event" to occur in this probability space: either player one picks up all 4 red balls (first type of good event), or player 1 picks up none of the red balls (second type of good event; equivalent to saying player 2 picks up all red balls).



                          The total number of ways that the first type of good event can occur is the number of ways we can pick 10 - 4 = 6 balls amond 24 - 4 = 20 balls after we fix that player one has selected the 4 red balls. In other words, there are $20 choose 6$ ways the first type of good event can occur.



                          The total number of ways that the second type of good event may occur is the number of ways player one can pick all of his balls exclusively from the white balls. Since there are 20 white balls and player 1 picks 10 balls in total, this gives us $20 choose 10$ ways for the second type of good event to occur.



                          Since the two "good" event spaces are mutually exclusive (player 1 can't pick all the red balls and none of the red balls at the same time), we can add them up obtain the total number of good event (the numerator in the solution you provided), and divide this by the size of the probability space (the total number of ways player 1 can pick 10 balls from 24, which is your denominator term).



                          Does this answer your question?






                          share|cite|improve this answer








                          New contributor




                          ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            There are two ways for a "good event" to occur in this probability space: either player one picks up all 4 red balls (first type of good event), or player 1 picks up none of the red balls (second type of good event; equivalent to saying player 2 picks up all red balls).



                            The total number of ways that the first type of good event can occur is the number of ways we can pick 10 - 4 = 6 balls amond 24 - 4 = 20 balls after we fix that player one has selected the 4 red balls. In other words, there are $20 choose 6$ ways the first type of good event can occur.



                            The total number of ways that the second type of good event may occur is the number of ways player one can pick all of his balls exclusively from the white balls. Since there are 20 white balls and player 1 picks 10 balls in total, this gives us $20 choose 10$ ways for the second type of good event to occur.



                            Since the two "good" event spaces are mutually exclusive (player 1 can't pick all the red balls and none of the red balls at the same time), we can add them up obtain the total number of good event (the numerator in the solution you provided), and divide this by the size of the probability space (the total number of ways player 1 can pick 10 balls from 24, which is your denominator term).



                            Does this answer your question?






                            share|cite|improve this answer








                            New contributor




                            ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$



                            There are two ways for a "good event" to occur in this probability space: either player one picks up all 4 red balls (first type of good event), or player 1 picks up none of the red balls (second type of good event; equivalent to saying player 2 picks up all red balls).



                            The total number of ways that the first type of good event can occur is the number of ways we can pick 10 - 4 = 6 balls amond 24 - 4 = 20 balls after we fix that player one has selected the 4 red balls. In other words, there are $20 choose 6$ ways the first type of good event can occur.



                            The total number of ways that the second type of good event may occur is the number of ways player one can pick all of his balls exclusively from the white balls. Since there are 20 white balls and player 1 picks 10 balls in total, this gives us $20 choose 10$ ways for the second type of good event to occur.



                            Since the two "good" event spaces are mutually exclusive (player 1 can't pick all the red balls and none of the red balls at the same time), we can add them up obtain the total number of good event (the numerator in the solution you provided), and divide this by the size of the probability space (the total number of ways player 1 can pick 10 balls from 24, which is your denominator term).



                            Does this answer your question?







                            share|cite|improve this answer








                            New contributor




                            ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|cite|improve this answer



                            share|cite|improve this answer






                            New contributor




                            ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered 2 days ago









                            ChrisHansenChrisHansen

                            111




                            111




                            New contributor




                            ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





                            New contributor





                            ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            ChrisHansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





















                                1












                                $begingroup$

                                Firstly, note that the events are mutually exclusive events. Therefore, the probability overall is the sum of the probability of the first person and second person selecting 4 red balls.



                                The number of ways the first person selects the 4 red balls is the same as the number of ways they can select the remaining six as white balls (they have 10 chances in total) from the 20 white balls available. This can be done in $24-4choose6$ ways



                                The number of ways the second person selects the 4 red balls is the same as the number of ways they can select the remaining ten as white balls (they have 14 chances on total) from the 20 white balls available. This can be done in $24-4choose10$ ways.



                                The total number of ways to split the 24 balls is $24choose10$



                                Therefore, the probability is what you put down.






                                share|cite|improve this answer










                                New contributor




                                aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$

















                                  1












                                  $begingroup$

                                  Firstly, note that the events are mutually exclusive events. Therefore, the probability overall is the sum of the probability of the first person and second person selecting 4 red balls.



                                  The number of ways the first person selects the 4 red balls is the same as the number of ways they can select the remaining six as white balls (they have 10 chances in total) from the 20 white balls available. This can be done in $24-4choose6$ ways



                                  The number of ways the second person selects the 4 red balls is the same as the number of ways they can select the remaining ten as white balls (they have 14 chances on total) from the 20 white balls available. This can be done in $24-4choose10$ ways.



                                  The total number of ways to split the 24 balls is $24choose10$



                                  Therefore, the probability is what you put down.






                                  share|cite|improve this answer










                                  New contributor




                                  aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.






                                  $endgroup$















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Firstly, note that the events are mutually exclusive events. Therefore, the probability overall is the sum of the probability of the first person and second person selecting 4 red balls.



                                    The number of ways the first person selects the 4 red balls is the same as the number of ways they can select the remaining six as white balls (they have 10 chances in total) from the 20 white balls available. This can be done in $24-4choose6$ ways



                                    The number of ways the second person selects the 4 red balls is the same as the number of ways they can select the remaining ten as white balls (they have 14 chances on total) from the 20 white balls available. This can be done in $24-4choose10$ ways.



                                    The total number of ways to split the 24 balls is $24choose10$



                                    Therefore, the probability is what you put down.






                                    share|cite|improve this answer










                                    New contributor




                                    aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    $endgroup$



                                    Firstly, note that the events are mutually exclusive events. Therefore, the probability overall is the sum of the probability of the first person and second person selecting 4 red balls.



                                    The number of ways the first person selects the 4 red balls is the same as the number of ways they can select the remaining six as white balls (they have 10 chances in total) from the 20 white balls available. This can be done in $24-4choose6$ ways



                                    The number of ways the second person selects the 4 red balls is the same as the number of ways they can select the remaining ten as white balls (they have 14 chances on total) from the 20 white balls available. This can be done in $24-4choose10$ ways.



                                    The total number of ways to split the 24 balls is $24choose10$



                                    Therefore, the probability is what you put down.







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