What's the probability to select socks of the same color?












2












$begingroup$


I'm sorry if this has been answered before already.



There's a drawer with 20 black socks and 20 white socks. We take one sock. What's the probability that the randomly chosen second sock has the same color as the second one?



I have two ways of thinking about it:




  1. The color of the first sock doesn't really matter, in any case, there're 19 socks of the same color left in the drawer and 39 in total. So that answer is $frac{19}{39}$.


  2. We have two cases: the first sock was black or it was white. The probability of taking white sock after white sock is equal to the probability of taking black sock after black sock and is equal to $frac{19}{39}$. The probability that the first sock is black is the same that it's white and is 0.5.
    In general case we can apply The Law of Total Probability:
    begin{equation}
    frac{19}{39} * 0.5 + frac{19}{39} * 0.5 = frac{19}{39}
    end{equation}



Does this reasoning make sense?










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$endgroup$












  • $begingroup$
    This is essentially the same answer twice, right both times.
    $endgroup$
    – Ethan Bolker
    Jan 12 at 18:11










  • $begingroup$
    @EthanBolker yes, the second would be more useful if the were 20 black and 10 white socks for example
    $endgroup$
    – mighter
    Jan 12 at 18:17
















2












$begingroup$


I'm sorry if this has been answered before already.



There's a drawer with 20 black socks and 20 white socks. We take one sock. What's the probability that the randomly chosen second sock has the same color as the second one?



I have two ways of thinking about it:




  1. The color of the first sock doesn't really matter, in any case, there're 19 socks of the same color left in the drawer and 39 in total. So that answer is $frac{19}{39}$.


  2. We have two cases: the first sock was black or it was white. The probability of taking white sock after white sock is equal to the probability of taking black sock after black sock and is equal to $frac{19}{39}$. The probability that the first sock is black is the same that it's white and is 0.5.
    In general case we can apply The Law of Total Probability:
    begin{equation}
    frac{19}{39} * 0.5 + frac{19}{39} * 0.5 = frac{19}{39}
    end{equation}



Does this reasoning make sense?










share|cite|improve this question









$endgroup$












  • $begingroup$
    This is essentially the same answer twice, right both times.
    $endgroup$
    – Ethan Bolker
    Jan 12 at 18:11










  • $begingroup$
    @EthanBolker yes, the second would be more useful if the were 20 black and 10 white socks for example
    $endgroup$
    – mighter
    Jan 12 at 18:17














2












2








2





$begingroup$


I'm sorry if this has been answered before already.



There's a drawer with 20 black socks and 20 white socks. We take one sock. What's the probability that the randomly chosen second sock has the same color as the second one?



I have two ways of thinking about it:




  1. The color of the first sock doesn't really matter, in any case, there're 19 socks of the same color left in the drawer and 39 in total. So that answer is $frac{19}{39}$.


  2. We have two cases: the first sock was black or it was white. The probability of taking white sock after white sock is equal to the probability of taking black sock after black sock and is equal to $frac{19}{39}$. The probability that the first sock is black is the same that it's white and is 0.5.
    In general case we can apply The Law of Total Probability:
    begin{equation}
    frac{19}{39} * 0.5 + frac{19}{39} * 0.5 = frac{19}{39}
    end{equation}



Does this reasoning make sense?










share|cite|improve this question









$endgroup$




I'm sorry if this has been answered before already.



There's a drawer with 20 black socks and 20 white socks. We take one sock. What's the probability that the randomly chosen second sock has the same color as the second one?



I have two ways of thinking about it:




  1. The color of the first sock doesn't really matter, in any case, there're 19 socks of the same color left in the drawer and 39 in total. So that answer is $frac{19}{39}$.


  2. We have two cases: the first sock was black or it was white. The probability of taking white sock after white sock is equal to the probability of taking black sock after black sock and is equal to $frac{19}{39}$. The probability that the first sock is black is the same that it's white and is 0.5.
    In general case we can apply The Law of Total Probability:
    begin{equation}
    frac{19}{39} * 0.5 + frac{19}{39} * 0.5 = frac{19}{39}
    end{equation}



Does this reasoning make sense?







probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 18:07









mightermighter

132




132












  • $begingroup$
    This is essentially the same answer twice, right both times.
    $endgroup$
    – Ethan Bolker
    Jan 12 at 18:11










  • $begingroup$
    @EthanBolker yes, the second would be more useful if the were 20 black and 10 white socks for example
    $endgroup$
    – mighter
    Jan 12 at 18:17


















  • $begingroup$
    This is essentially the same answer twice, right both times.
    $endgroup$
    – Ethan Bolker
    Jan 12 at 18:11










  • $begingroup$
    @EthanBolker yes, the second would be more useful if the were 20 black and 10 white socks for example
    $endgroup$
    – mighter
    Jan 12 at 18:17
















$begingroup$
This is essentially the same answer twice, right both times.
$endgroup$
– Ethan Bolker
Jan 12 at 18:11




$begingroup$
This is essentially the same answer twice, right both times.
$endgroup$
– Ethan Bolker
Jan 12 at 18:11












$begingroup$
@EthanBolker yes, the second would be more useful if the were 20 black and 10 white socks for example
$endgroup$
– mighter
Jan 12 at 18:17




$begingroup$
@EthanBolker yes, the second would be more useful if the were 20 black and 10 white socks for example
$endgroup$
– mighter
Jan 12 at 18:17










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let us denote by $B_{k}$ the event "a black sock has been chosen in the $k$-th selection". The same applies to $W_{k}$. Thus we are interested in the event $(B_{1}cap B_{2})cup(W_{1}cap W_{2})$, where $B_{1}cap B_{2}$ and $W_{1}cap W_{2}$ are disjoint. Therefore we have
begin{align*}
textbf{P}((B_{1}cap B_{2})cup(W_{1}cap W_{2})) & = textbf{P}(B_{1}cap B_{2}) + textbf{P}(W_{1}cap W_{2})\
& = textbf{P}(B_{2}|B_{1})textbf{P}(B_{1}) + textbf{P}(W_{2}|W_{1})textbf{P}(W_{1})\
& = frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)} + frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)}\
& = frac{19}{39}timesfrac{1}{2} + frac{19}{39}timesfrac{1}{2} = frac{19}{39}
end{align*}



This is a more "systematic" way to describe the problem. Hope this helps.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Both the reasonings are correct. Consider the case when all balls socks $s_1,dots, s_{40}$ are of different colors. First, the probability of selecting a sock of a particular color is $frac{1}{40}$. Second, the probability of selecting a sock of a particular color from the rest of the 39 socks is $frac{1}{39}$.



    Now apply counting principle. Since in $s_1,dots, s_{40}$ there are 20 of the same color (black), the probability of the first event is $20 times frac{1}{40}=frac{1}{2}$. Thus the probability of getting a black sock in the first event is $frac{1}{2}$, similarly, the probability of getting a white sock in the first event is $frac{1}{2}$.



    For the second event, you have two cases, either the sock in the first event is black or white. Thus the probability is $2timesfrac{1}{2}times frac{19}{40}=frac{19}{40}$. The only difference is that your first answer is not using the counting principal whereas the second answer does.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Could you explain more?
      $endgroup$
      – Larry
      Jan 12 at 18:52











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    0












    $begingroup$

    Let us denote by $B_{k}$ the event "a black sock has been chosen in the $k$-th selection". The same applies to $W_{k}$. Thus we are interested in the event $(B_{1}cap B_{2})cup(W_{1}cap W_{2})$, where $B_{1}cap B_{2}$ and $W_{1}cap W_{2}$ are disjoint. Therefore we have
    begin{align*}
    textbf{P}((B_{1}cap B_{2})cup(W_{1}cap W_{2})) & = textbf{P}(B_{1}cap B_{2}) + textbf{P}(W_{1}cap W_{2})\
    & = textbf{P}(B_{2}|B_{1})textbf{P}(B_{1}) + textbf{P}(W_{2}|W_{1})textbf{P}(W_{1})\
    & = frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)} + frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)}\
    & = frac{19}{39}timesfrac{1}{2} + frac{19}{39}timesfrac{1}{2} = frac{19}{39}
    end{align*}



    This is a more "systematic" way to describe the problem. Hope this helps.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let us denote by $B_{k}$ the event "a black sock has been chosen in the $k$-th selection". The same applies to $W_{k}$. Thus we are interested in the event $(B_{1}cap B_{2})cup(W_{1}cap W_{2})$, where $B_{1}cap B_{2}$ and $W_{1}cap W_{2}$ are disjoint. Therefore we have
      begin{align*}
      textbf{P}((B_{1}cap B_{2})cup(W_{1}cap W_{2})) & = textbf{P}(B_{1}cap B_{2}) + textbf{P}(W_{1}cap W_{2})\
      & = textbf{P}(B_{2}|B_{1})textbf{P}(B_{1}) + textbf{P}(W_{2}|W_{1})textbf{P}(W_{1})\
      & = frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)} + frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)}\
      & = frac{19}{39}timesfrac{1}{2} + frac{19}{39}timesfrac{1}{2} = frac{19}{39}
      end{align*}



      This is a more "systematic" way to describe the problem. Hope this helps.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let us denote by $B_{k}$ the event "a black sock has been chosen in the $k$-th selection". The same applies to $W_{k}$. Thus we are interested in the event $(B_{1}cap B_{2})cup(W_{1}cap W_{2})$, where $B_{1}cap B_{2}$ and $W_{1}cap W_{2}$ are disjoint. Therefore we have
        begin{align*}
        textbf{P}((B_{1}cap B_{2})cup(W_{1}cap W_{2})) & = textbf{P}(B_{1}cap B_{2}) + textbf{P}(W_{1}cap W_{2})\
        & = textbf{P}(B_{2}|B_{1})textbf{P}(B_{1}) + textbf{P}(W_{2}|W_{1})textbf{P}(W_{1})\
        & = frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)} + frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)}\
        & = frac{19}{39}timesfrac{1}{2} + frac{19}{39}timesfrac{1}{2} = frac{19}{39}
        end{align*}



        This is a more "systematic" way to describe the problem. Hope this helps.






        share|cite|improve this answer











        $endgroup$



        Let us denote by $B_{k}$ the event "a black sock has been chosen in the $k$-th selection". The same applies to $W_{k}$. Thus we are interested in the event $(B_{1}cap B_{2})cup(W_{1}cap W_{2})$, where $B_{1}cap B_{2}$ and $W_{1}cap W_{2}$ are disjoint. Therefore we have
        begin{align*}
        textbf{P}((B_{1}cap B_{2})cup(W_{1}cap W_{2})) & = textbf{P}(B_{1}cap B_{2}) + textbf{P}(W_{1}cap W_{2})\
        & = textbf{P}(B_{2}|B_{1})textbf{P}(B_{1}) + textbf{P}(W_{2}|W_{1})textbf{P}(W_{1})\
        & = frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)} + frac{C(19,1)}{C(39,1)}timesfrac{C(20,1)}{C(40,1)}\
        & = frac{19}{39}timesfrac{1}{2} + frac{19}{39}timesfrac{1}{2} = frac{19}{39}
        end{align*}



        This is a more "systematic" way to describe the problem. Hope this helps.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 18:25

























        answered Jan 12 at 18:18









        user1337user1337

        43010




        43010























            0












            $begingroup$

            Both the reasonings are correct. Consider the case when all balls socks $s_1,dots, s_{40}$ are of different colors. First, the probability of selecting a sock of a particular color is $frac{1}{40}$. Second, the probability of selecting a sock of a particular color from the rest of the 39 socks is $frac{1}{39}$.



            Now apply counting principle. Since in $s_1,dots, s_{40}$ there are 20 of the same color (black), the probability of the first event is $20 times frac{1}{40}=frac{1}{2}$. Thus the probability of getting a black sock in the first event is $frac{1}{2}$, similarly, the probability of getting a white sock in the first event is $frac{1}{2}$.



            For the second event, you have two cases, either the sock in the first event is black or white. Thus the probability is $2timesfrac{1}{2}times frac{19}{40}=frac{19}{40}$. The only difference is that your first answer is not using the counting principal whereas the second answer does.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Could you explain more?
              $endgroup$
              – Larry
              Jan 12 at 18:52
















            0












            $begingroup$

            Both the reasonings are correct. Consider the case when all balls socks $s_1,dots, s_{40}$ are of different colors. First, the probability of selecting a sock of a particular color is $frac{1}{40}$. Second, the probability of selecting a sock of a particular color from the rest of the 39 socks is $frac{1}{39}$.



            Now apply counting principle. Since in $s_1,dots, s_{40}$ there are 20 of the same color (black), the probability of the first event is $20 times frac{1}{40}=frac{1}{2}$. Thus the probability of getting a black sock in the first event is $frac{1}{2}$, similarly, the probability of getting a white sock in the first event is $frac{1}{2}$.



            For the second event, you have two cases, either the sock in the first event is black or white. Thus the probability is $2timesfrac{1}{2}times frac{19}{40}=frac{19}{40}$. The only difference is that your first answer is not using the counting principal whereas the second answer does.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Could you explain more?
              $endgroup$
              – Larry
              Jan 12 at 18:52














            0












            0








            0





            $begingroup$

            Both the reasonings are correct. Consider the case when all balls socks $s_1,dots, s_{40}$ are of different colors. First, the probability of selecting a sock of a particular color is $frac{1}{40}$. Second, the probability of selecting a sock of a particular color from the rest of the 39 socks is $frac{1}{39}$.



            Now apply counting principle. Since in $s_1,dots, s_{40}$ there are 20 of the same color (black), the probability of the first event is $20 times frac{1}{40}=frac{1}{2}$. Thus the probability of getting a black sock in the first event is $frac{1}{2}$, similarly, the probability of getting a white sock in the first event is $frac{1}{2}$.



            For the second event, you have two cases, either the sock in the first event is black or white. Thus the probability is $2timesfrac{1}{2}times frac{19}{40}=frac{19}{40}$. The only difference is that your first answer is not using the counting principal whereas the second answer does.






            share|cite|improve this answer











            $endgroup$



            Both the reasonings are correct. Consider the case when all balls socks $s_1,dots, s_{40}$ are of different colors. First, the probability of selecting a sock of a particular color is $frac{1}{40}$. Second, the probability of selecting a sock of a particular color from the rest of the 39 socks is $frac{1}{39}$.



            Now apply counting principle. Since in $s_1,dots, s_{40}$ there are 20 of the same color (black), the probability of the first event is $20 times frac{1}{40}=frac{1}{2}$. Thus the probability of getting a black sock in the first event is $frac{1}{2}$, similarly, the probability of getting a white sock in the first event is $frac{1}{2}$.



            For the second event, you have two cases, either the sock in the first event is black or white. Thus the probability is $2timesfrac{1}{2}times frac{19}{40}=frac{19}{40}$. The only difference is that your first answer is not using the counting principal whereas the second answer does.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 13 at 5:44

























            answered Jan 12 at 18:12









            Ranveer SinghRanveer Singh

            209210




            209210












            • $begingroup$
              Could you explain more?
              $endgroup$
              – Larry
              Jan 12 at 18:52


















            • $begingroup$
              Could you explain more?
              $endgroup$
              – Larry
              Jan 12 at 18:52
















            $begingroup$
            Could you explain more?
            $endgroup$
            – Larry
            Jan 12 at 18:52




            $begingroup$
            Could you explain more?
            $endgroup$
            – Larry
            Jan 12 at 18:52


















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