probability - selecting balls from a box












0












$begingroup$


A box contains 4 white, 5 red, and 6 black balls, four balls are drawn at random from the box. Then the probability that among the balls drawn there is at least 1 ball of each color is?



What goes wrong in my approach:
$$frac{4C1 cdot 5C1 cdot 6C1 cdot 12}{15C4}$$






probability combinatorics combinations







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$endgroup$








  • 2




    $begingroup$
    You are overcounting. You count the choice $(W_1,R_1,B_1,W_2)$ and $(W_2,R_1,B_1,W_1)$ as two distinct choices.
    $endgroup$
    – lulu
    Dec 17 '18 at 15:09






  • 1




    $begingroup$
    oh got it. Thanks I should have done selecting 2 balls from each color followed by other 2 colors one ball each.
    $endgroup$
    – swapnil
    Dec 17 '18 at 15:13






  • 1




    $begingroup$
    As an alternative: you can argue that your method exactly double counts. That way, all you need to do is to divide your answer by $2$.
    $endgroup$
    – lulu
    Dec 17 '18 at 15:19










  • $begingroup$
    I would do it like that : $p = frac{3 cdot 4!}{15C4}$ where the nominator describe the permutations of 4 balls, 3 of different colors and the last one is either red or white or black
    $endgroup$
    – Alain.Klbtr
    Dec 17 '18 at 15:32












  • $begingroup$
    This tutorial explains how to typeset mathematics on this site. Also, when you pose an attempt here, you should explain how you got your answer since that helps users identify any errors you may have made.
    $endgroup$
    – N. F. Taussig
    Dec 17 '18 at 15:32


















0












$begingroup$


A box contains 4 white, 5 red, and 6 black balls, four balls are drawn at random from the box. Then the probability that among the balls drawn there is at least 1 ball of each color is?



What goes wrong in my approach:
$$frac{4C1 cdot 5C1 cdot 6C1 cdot 12}{15C4}$$






probability combinatorics combinations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You are overcounting. You count the choice $(W_1,R_1,B_1,W_2)$ and $(W_2,R_1,B_1,W_1)$ as two distinct choices.
    $endgroup$
    – lulu
    Dec 17 '18 at 15:09






  • 1




    $begingroup$
    oh got it. Thanks I should have done selecting 2 balls from each color followed by other 2 colors one ball each.
    $endgroup$
    – swapnil
    Dec 17 '18 at 15:13






  • 1




    $begingroup$
    As an alternative: you can argue that your method exactly double counts. That way, all you need to do is to divide your answer by $2$.
    $endgroup$
    – lulu
    Dec 17 '18 at 15:19










  • $begingroup$
    I would do it like that : $p = frac{3 cdot 4!}{15C4}$ where the nominator describe the permutations of 4 balls, 3 of different colors and the last one is either red or white or black
    $endgroup$
    – Alain.Klbtr
    Dec 17 '18 at 15:32












  • $begingroup$
    This tutorial explains how to typeset mathematics on this site. Also, when you pose an attempt here, you should explain how you got your answer since that helps users identify any errors you may have made.
    $endgroup$
    – N. F. Taussig
    Dec 17 '18 at 15:32
















0












0








0





$begingroup$


A box contains 4 white, 5 red, and 6 black balls, four balls are drawn at random from the box. Then the probability that among the balls drawn there is at least 1 ball of each color is?



What goes wrong in my approach:
$$frac{4C1 cdot 5C1 cdot 6C1 cdot 12}{15C4}$$






probability combinatorics combinations







share|cite|improve this question











$endgroup$




A box contains 4 white, 5 red, and 6 black balls, four balls are drawn at random from the box. Then the probability that among the balls drawn there is at least 1 ball of each color is?



What goes wrong in my approach:
$$frac{4C1 cdot 5C1 cdot 6C1 cdot 12}{15C4}$$






probability combinatorics combinations




probability combinatorics combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 16:18









N. F. Taussig

44.3k93357




44.3k93357










asked Dec 17 '18 at 15:06









swapnilswapnil

335




335








  • 2




    $begingroup$
    You are overcounting. You count the choice $(W_1,R_1,B_1,W_2)$ and $(W_2,R_1,B_1,W_1)$ as two distinct choices.
    $endgroup$
    – lulu
    Dec 17 '18 at 15:09






  • 1




    $begingroup$
    oh got it. Thanks I should have done selecting 2 balls from each color followed by other 2 colors one ball each.
    $endgroup$
    – swapnil
    Dec 17 '18 at 15:13






  • 1




    $begingroup$
    As an alternative: you can argue that your method exactly double counts. That way, all you need to do is to divide your answer by $2$.
    $endgroup$
    – lulu
    Dec 17 '18 at 15:19










  • $begingroup$
    I would do it like that : $p = frac{3 cdot 4!}{15C4}$ where the nominator describe the permutations of 4 balls, 3 of different colors and the last one is either red or white or black
    $endgroup$
    – Alain.Klbtr
    Dec 17 '18 at 15:32












  • $begingroup$
    This tutorial explains how to typeset mathematics on this site. Also, when you pose an attempt here, you should explain how you got your answer since that helps users identify any errors you may have made.
    $endgroup$
    – N. F. Taussig
    Dec 17 '18 at 15:32
















  • 2




    $begingroup$
    You are overcounting. You count the choice $(W_1,R_1,B_1,W_2)$ and $(W_2,R_1,B_1,W_1)$ as two distinct choices.
    $endgroup$
    – lulu
    Dec 17 '18 at 15:09






  • 1




    $begingroup$
    oh got it. Thanks I should have done selecting 2 balls from each color followed by other 2 colors one ball each.
    $endgroup$
    – swapnil
    Dec 17 '18 at 15:13






  • 1




    $begingroup$
    As an alternative: you can argue that your method exactly double counts. That way, all you need to do is to divide your answer by $2$.
    $endgroup$
    – lulu
    Dec 17 '18 at 15:19










  • $begingroup$
    I would do it like that : $p = frac{3 cdot 4!}{15C4}$ where the nominator describe the permutations of 4 balls, 3 of different colors and the last one is either red or white or black
    $endgroup$
    – Alain.Klbtr
    Dec 17 '18 at 15:32












  • $begingroup$
    This tutorial explains how to typeset mathematics on this site. Also, when you pose an attempt here, you should explain how you got your answer since that helps users identify any errors you may have made.
    $endgroup$
    – N. F. Taussig
    Dec 17 '18 at 15:32










2




2




$begingroup$
You are overcounting. You count the choice $(W_1,R_1,B_1,W_2)$ and $(W_2,R_1,B_1,W_1)$ as two distinct choices.
$endgroup$
– lulu
Dec 17 '18 at 15:09




$begingroup$
You are overcounting. You count the choice $(W_1,R_1,B_1,W_2)$ and $(W_2,R_1,B_1,W_1)$ as two distinct choices.
$endgroup$
– lulu
Dec 17 '18 at 15:09




1




1




$begingroup$
oh got it. Thanks I should have done selecting 2 balls from each color followed by other 2 colors one ball each.
$endgroup$
– swapnil
Dec 17 '18 at 15:13




$begingroup$
oh got it. Thanks I should have done selecting 2 balls from each color followed by other 2 colors one ball each.
$endgroup$
– swapnil
Dec 17 '18 at 15:13




1




1




$begingroup$
As an alternative: you can argue that your method exactly double counts. That way, all you need to do is to divide your answer by $2$.
$endgroup$
– lulu
Dec 17 '18 at 15:19




$begingroup$
As an alternative: you can argue that your method exactly double counts. That way, all you need to do is to divide your answer by $2$.
$endgroup$
– lulu
Dec 17 '18 at 15:19












$begingroup$
I would do it like that : $p = frac{3 cdot 4!}{15C4}$ where the nominator describe the permutations of 4 balls, 3 of different colors and the last one is either red or white or black
$endgroup$
– Alain.Klbtr
Dec 17 '18 at 15:32






$begingroup$
I would do it like that : $p = frac{3 cdot 4!}{15C4}$ where the nominator describe the permutations of 4 balls, 3 of different colors and the last one is either red or white or black
$endgroup$
– Alain.Klbtr
Dec 17 '18 at 15:32














$begingroup$
This tutorial explains how to typeset mathematics on this site. Also, when you pose an attempt here, you should explain how you got your answer since that helps users identify any errors you may have made.
$endgroup$
– N. F. Taussig
Dec 17 '18 at 15:32






$begingroup$
This tutorial explains how to typeset mathematics on this site. Also, when you pose an attempt here, you should explain how you got your answer since that helps users identify any errors you may have made.
$endgroup$
– N. F. Taussig
Dec 17 '18 at 15:32












1 Answer
1






active

oldest

votes


















1












$begingroup$

Your method counts every favorable selection twice, once for each way you could designate one of the balls from the color that is selected twice as the ball of that color.



Let's compare your approach with the correct solution.



Solution: The number of ways we may select four of the $4 + 5 + 6 = 15$ balls is
$$binom{15}{4}$$



If at least one ball of each color is selected, then since four balls are being selected from three colors, exactly one color must be selected twice with the other two colors each selected once.



Two white balls, one red ball, and one black ball are selected: We must select two of the four white balls, one of the five red balls, and one of the six black balls, which can be done in
$$binom{4}{2}binom{5}{1}binom{6}{1}$$
ways.



One white ball, two red balls, and one black ball are selected: We must select one of the four white balls, two of the five red balls, and one of the six black balls, which can be done in
$$binom{4}{1}binom{5}{2}binom{6}{1}$$
ways.



One white ball, one red ball, and two black balls are selected: We must select one of the four white balls, one of the five red balls, and two of the six black balls, which can be done in
$$binom{4}{1}binom{5}{1}binom{6}{2}$$
ways.



Number of favorable cases: Since the three cases above are mutually exclusive and exhaustive, the number of favorable cases is found by adding the number of selections for each case.
$$binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}binom{6}{2}$$



Probability: Assuming each ball is selected randomly, the probability of selecting at least one ball of each color is
$$frac{dbinom{4}{2}dbinom{5}{1}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{2}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{1}dbinom{6}{2}}{dbinom{15}{4}}$$



What error did you make?



By designating a ball of each color as the ball reserved for that color, you count each favorable selection twice, once for each way you could have designated one of those balls as the ball of that color. To borrow @lulu's example, you count the selection $W_1, W_2, R_1, B_1$ twice:



begin{array}{c c}
text{designated balls} & text{additional ball}\ hline
W_1, R_1, B_1 & W_2\
W_2, R_1, B_1 & W_1
end{array}



Notice that
$$color{red}{binom{2}{1}}binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}color{red}{binom{2}{1}}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}color{red}{binom{2}{1}}binom{6}{2} = binom{4}{1}binom{5}{1}binom{6}{1} cdot 12$$
where the factor $color{red}{binom{2}{1}}$ represents the number of ways you could have designated one of the two balls from the color that is selected twice as the representative ball of that color.






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    oldest

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    1












    $begingroup$

    Your method counts every favorable selection twice, once for each way you could designate one of the balls from the color that is selected twice as the ball of that color.



    Let's compare your approach with the correct solution.



    Solution: The number of ways we may select four of the $4 + 5 + 6 = 15$ balls is
    $$binom{15}{4}$$



    If at least one ball of each color is selected, then since four balls are being selected from three colors, exactly one color must be selected twice with the other two colors each selected once.



    Two white balls, one red ball, and one black ball are selected: We must select two of the four white balls, one of the five red balls, and one of the six black balls, which can be done in
    $$binom{4}{2}binom{5}{1}binom{6}{1}$$
    ways.



    One white ball, two red balls, and one black ball are selected: We must select one of the four white balls, two of the five red balls, and one of the six black balls, which can be done in
    $$binom{4}{1}binom{5}{2}binom{6}{1}$$
    ways.



    One white ball, one red ball, and two black balls are selected: We must select one of the four white balls, one of the five red balls, and two of the six black balls, which can be done in
    $$binom{4}{1}binom{5}{1}binom{6}{2}$$
    ways.



    Number of favorable cases: Since the three cases above are mutually exclusive and exhaustive, the number of favorable cases is found by adding the number of selections for each case.
    $$binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}binom{6}{2}$$



    Probability: Assuming each ball is selected randomly, the probability of selecting at least one ball of each color is
    $$frac{dbinom{4}{2}dbinom{5}{1}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{2}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{1}dbinom{6}{2}}{dbinom{15}{4}}$$



    What error did you make?



    By designating a ball of each color as the ball reserved for that color, you count each favorable selection twice, once for each way you could have designated one of those balls as the ball of that color. To borrow @lulu's example, you count the selection $W_1, W_2, R_1, B_1$ twice:



    begin{array}{c c}
    text{designated balls} & text{additional ball}\ hline
    W_1, R_1, B_1 & W_2\
    W_2, R_1, B_1 & W_1
    end{array}



    Notice that
    $$color{red}{binom{2}{1}}binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}color{red}{binom{2}{1}}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}color{red}{binom{2}{1}}binom{6}{2} = binom{4}{1}binom{5}{1}binom{6}{1} cdot 12$$
    where the factor $color{red}{binom{2}{1}}$ represents the number of ways you could have designated one of the two balls from the color that is selected twice as the representative ball of that color.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Your method counts every favorable selection twice, once for each way you could designate one of the balls from the color that is selected twice as the ball of that color.



      Let's compare your approach with the correct solution.



      Solution: The number of ways we may select four of the $4 + 5 + 6 = 15$ balls is
      $$binom{15}{4}$$



      If at least one ball of each color is selected, then since four balls are being selected from three colors, exactly one color must be selected twice with the other two colors each selected once.



      Two white balls, one red ball, and one black ball are selected: We must select two of the four white balls, one of the five red balls, and one of the six black balls, which can be done in
      $$binom{4}{2}binom{5}{1}binom{6}{1}$$
      ways.



      One white ball, two red balls, and one black ball are selected: We must select one of the four white balls, two of the five red balls, and one of the six black balls, which can be done in
      $$binom{4}{1}binom{5}{2}binom{6}{1}$$
      ways.



      One white ball, one red ball, and two black balls are selected: We must select one of the four white balls, one of the five red balls, and two of the six black balls, which can be done in
      $$binom{4}{1}binom{5}{1}binom{6}{2}$$
      ways.



      Number of favorable cases: Since the three cases above are mutually exclusive and exhaustive, the number of favorable cases is found by adding the number of selections for each case.
      $$binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}binom{6}{2}$$



      Probability: Assuming each ball is selected randomly, the probability of selecting at least one ball of each color is
      $$frac{dbinom{4}{2}dbinom{5}{1}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{2}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{1}dbinom{6}{2}}{dbinom{15}{4}}$$



      What error did you make?



      By designating a ball of each color as the ball reserved for that color, you count each favorable selection twice, once for each way you could have designated one of those balls as the ball of that color. To borrow @lulu's example, you count the selection $W_1, W_2, R_1, B_1$ twice:



      begin{array}{c c}
      text{designated balls} & text{additional ball}\ hline
      W_1, R_1, B_1 & W_2\
      W_2, R_1, B_1 & W_1
      end{array}



      Notice that
      $$color{red}{binom{2}{1}}binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}color{red}{binom{2}{1}}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}color{red}{binom{2}{1}}binom{6}{2} = binom{4}{1}binom{5}{1}binom{6}{1} cdot 12$$
      where the factor $color{red}{binom{2}{1}}$ represents the number of ways you could have designated one of the two balls from the color that is selected twice as the representative ball of that color.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Your method counts every favorable selection twice, once for each way you could designate one of the balls from the color that is selected twice as the ball of that color.



        Let's compare your approach with the correct solution.



        Solution: The number of ways we may select four of the $4 + 5 + 6 = 15$ balls is
        $$binom{15}{4}$$



        If at least one ball of each color is selected, then since four balls are being selected from three colors, exactly one color must be selected twice with the other two colors each selected once.



        Two white balls, one red ball, and one black ball are selected: We must select two of the four white balls, one of the five red balls, and one of the six black balls, which can be done in
        $$binom{4}{2}binom{5}{1}binom{6}{1}$$
        ways.



        One white ball, two red balls, and one black ball are selected: We must select one of the four white balls, two of the five red balls, and one of the six black balls, which can be done in
        $$binom{4}{1}binom{5}{2}binom{6}{1}$$
        ways.



        One white ball, one red ball, and two black balls are selected: We must select one of the four white balls, one of the five red balls, and two of the six black balls, which can be done in
        $$binom{4}{1}binom{5}{1}binom{6}{2}$$
        ways.



        Number of favorable cases: Since the three cases above are mutually exclusive and exhaustive, the number of favorable cases is found by adding the number of selections for each case.
        $$binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}binom{6}{2}$$



        Probability: Assuming each ball is selected randomly, the probability of selecting at least one ball of each color is
        $$frac{dbinom{4}{2}dbinom{5}{1}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{2}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{1}dbinom{6}{2}}{dbinom{15}{4}}$$



        What error did you make?



        By designating a ball of each color as the ball reserved for that color, you count each favorable selection twice, once for each way you could have designated one of those balls as the ball of that color. To borrow @lulu's example, you count the selection $W_1, W_2, R_1, B_1$ twice:



        begin{array}{c c}
        text{designated balls} & text{additional ball}\ hline
        W_1, R_1, B_1 & W_2\
        W_2, R_1, B_1 & W_1
        end{array}



        Notice that
        $$color{red}{binom{2}{1}}binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}color{red}{binom{2}{1}}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}color{red}{binom{2}{1}}binom{6}{2} = binom{4}{1}binom{5}{1}binom{6}{1} cdot 12$$
        where the factor $color{red}{binom{2}{1}}$ represents the number of ways you could have designated one of the two balls from the color that is selected twice as the representative ball of that color.






        share|cite|improve this answer









        $endgroup$



        Your method counts every favorable selection twice, once for each way you could designate one of the balls from the color that is selected twice as the ball of that color.



        Let's compare your approach with the correct solution.



        Solution: The number of ways we may select four of the $4 + 5 + 6 = 15$ balls is
        $$binom{15}{4}$$



        If at least one ball of each color is selected, then since four balls are being selected from three colors, exactly one color must be selected twice with the other two colors each selected once.



        Two white balls, one red ball, and one black ball are selected: We must select two of the four white balls, one of the five red balls, and one of the six black balls, which can be done in
        $$binom{4}{2}binom{5}{1}binom{6}{1}$$
        ways.



        One white ball, two red balls, and one black ball are selected: We must select one of the four white balls, two of the five red balls, and one of the six black balls, which can be done in
        $$binom{4}{1}binom{5}{2}binom{6}{1}$$
        ways.



        One white ball, one red ball, and two black balls are selected: We must select one of the four white balls, one of the five red balls, and two of the six black balls, which can be done in
        $$binom{4}{1}binom{5}{1}binom{6}{2}$$
        ways.



        Number of favorable cases: Since the three cases above are mutually exclusive and exhaustive, the number of favorable cases is found by adding the number of selections for each case.
        $$binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}binom{6}{2}$$



        Probability: Assuming each ball is selected randomly, the probability of selecting at least one ball of each color is
        $$frac{dbinom{4}{2}dbinom{5}{1}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{2}dbinom{6}{1} + dbinom{4}{1}dbinom{5}{1}dbinom{6}{2}}{dbinom{15}{4}}$$



        What error did you make?



        By designating a ball of each color as the ball reserved for that color, you count each favorable selection twice, once for each way you could have designated one of those balls as the ball of that color. To borrow @lulu's example, you count the selection $W_1, W_2, R_1, B_1$ twice:



        begin{array}{c c}
        text{designated balls} & text{additional ball}\ hline
        W_1, R_1, B_1 & W_2\
        W_2, R_1, B_1 & W_1
        end{array}



        Notice that
        $$color{red}{binom{2}{1}}binom{4}{2}binom{5}{1}binom{6}{1} + binom{4}{1}color{red}{binom{2}{1}}binom{5}{2}binom{6}{1} + binom{4}{1}binom{5}{1}color{red}{binom{2}{1}}binom{6}{2} = binom{4}{1}binom{5}{1}binom{6}{1} cdot 12$$
        where the factor $color{red}{binom{2}{1}}$ represents the number of ways you could have designated one of the two balls from the color that is selected twice as the representative ball of that color.







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        answered Dec 18 '18 at 11:08









        N. F. TaussigN. F. Taussig

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