52 cards are equally given to 4 players Find probability that one of them has 3 spades out of remaining 5.












1












$begingroup$


52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.



In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ frac{binom{5}{3}binom{39}{10}}{binom{44}{13}},$$
is this approach correct?






probability







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$endgroup$












  • $begingroup$
    Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
    $endgroup$
    – lulu
    Dec 26 '15 at 4:11












  • $begingroup$
    Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
    $endgroup$
    – lulu
    Dec 26 '15 at 4:19


















1












$begingroup$


52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.



In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ frac{binom{5}{3}binom{39}{10}}{binom{44}{13}},$$
is this approach correct?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
    $endgroup$
    – lulu
    Dec 26 '15 at 4:11












  • $begingroup$
    Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
    $endgroup$
    – lulu
    Dec 26 '15 at 4:19
















1












1








1


2



$begingroup$


52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.



In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ frac{binom{5}{3}binom{39}{10}}{binom{44}{13}},$$
is this approach correct?






probability







share|cite|improve this question











$endgroup$




52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.



In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ frac{binom{5}{3}binom{39}{10}}{binom{44}{13}},$$
is this approach correct?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '15 at 4:21









Alex Provost

15.6k42351




15.6k42351










asked Dec 26 '15 at 4:06









radhikaradhika

141313




141313












  • $begingroup$
    Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
    $endgroup$
    – lulu
    Dec 26 '15 at 4:11












  • $begingroup$
    Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
    $endgroup$
    – lulu
    Dec 26 '15 at 4:19




















  • $begingroup$
    Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
    $endgroup$
    – lulu
    Dec 26 '15 at 4:11












  • $begingroup$
    Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
    $endgroup$
    – lulu
    Dec 26 '15 at 4:19


















$begingroup$
Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
$endgroup$
– lulu
Dec 26 '15 at 4:11






$begingroup$
Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
$endgroup$
– lulu
Dec 26 '15 at 4:11














$begingroup$
Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
$endgroup$
– lulu
Dec 26 '15 at 4:19






$begingroup$
Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
$endgroup$
– lulu
Dec 26 '15 at 4:19












3 Answers
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1












$begingroup$

There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?



    My Approach: My answer to the above question was:



    East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.



    All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.



    Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$



    However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      It is given that A and B have a total of 8 spades among 26 cards.
      ∴ In the remaining 26 cards, there are exactly 5 spades.
      These 26 cards are distributed equally among C and D [13 each]



      P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
      = 0.339






      share|cite|improve this answer









      $endgroup$














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        3 Answers
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        3 Answers
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        $begingroup$

        There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.






            share|cite|improve this answer











            $endgroup$



            There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 26 '15 at 5:14

























            answered Dec 26 '15 at 4:35









            MaffredMaffred

            2,675625




            2,675625























                1












                $begingroup$

                Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?



                My Approach: My answer to the above question was:



                East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.



                All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.



                Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$



                However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?



                  My Approach: My answer to the above question was:



                  East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.



                  All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.



                  Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$



                  However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?



                    My Approach: My answer to the above question was:



                    East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.



                    All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.



                    Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$



                    However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?






                    share|cite|improve this answer









                    $endgroup$



                    Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?



                    My Approach: My answer to the above question was:



                    East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.



                    All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.



                    Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$



                    However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jul 9 '16 at 20:11









                    Venkatesh VinayakaraoVenkatesh Vinayakarao

                    4316




                    4316























                        0












                        $begingroup$

                        It is given that A and B have a total of 8 spades among 26 cards.
                        ∴ In the remaining 26 cards, there are exactly 5 spades.
                        These 26 cards are distributed equally among C and D [13 each]



                        P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
                        = 0.339






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          It is given that A and B have a total of 8 spades among 26 cards.
                          ∴ In the remaining 26 cards, there are exactly 5 spades.
                          These 26 cards are distributed equally among C and D [13 each]



                          P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
                          = 0.339






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            It is given that A and B have a total of 8 spades among 26 cards.
                            ∴ In the remaining 26 cards, there are exactly 5 spades.
                            These 26 cards are distributed equally among C and D [13 each]



                            P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
                            = 0.339






                            share|cite|improve this answer









                            $endgroup$



                            It is given that A and B have a total of 8 spades among 26 cards.
                            ∴ In the remaining 26 cards, there are exactly 5 spades.
                            These 26 cards are distributed equally among C and D [13 each]



                            P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
                            = 0.339







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 23 '18 at 17:45









                            shivam guptashivam gupta

                            1




                            1






























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