<PRE><B>Question 1207921</B>
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A family of 6 boys and 4 girls. Mother wanted to choose 4 of them to prepare dinner:
What is the probability of choosing a boy to prepare tea, a boy to cook, and two girls to process
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&lt;pre&gt;
The total number of ways to assign jobs  {{{highlight(randomly)}}}  is

    {{{10*9*((8*7)/2)}}} = 10*9*28 = 2520.


(10 ways to fill the first position; 9 ways to fill the second position;
and {{{(8*7)/2)}}} = 28 ways to fill the 3-rd and 4-th positions with possible transposition there).


    +-------------------------------------------+
    |   Now let&#39;s calculate the number of ways  |
    |      to assign jobs as described.         |
    +-------------------------------------------+


The number of ways to select one boy from 6 boys to prepare tea is 6.            

The number of ways to select 1 boy from remaining 5 boys to cook is 5.

The number of ways to select 2 girls from 4 girls is  {{{C[4]^2}}} = {{{(4*3)/(1*2)}}} = 6.


So, the number of ways to form a proper team as described is 6*5*6 = 180.


The probability under the problem&#39;s question is

          number of proper ways      180     1
    P = ------------------------- = ----- = ---- = 0.07143  (rounded).    &lt;U&gt;ANSWER&lt;/U&gt;
          total number of ways       2520    14



    +---------------------------------------+
    |   &lt;U&gt;Another way&lt;/U&gt;  to solve is to write   |
    +---------------------------------------+


    P = {{{(6/10)*(5/9)*(C[4]^2/C[8]^2)}}} = {{{(3/5)*(5/9)*(6/28)}}} = {{{1/14}}}  (the same answer).


Here  {{{6/10}}}  is the probability to select one of 6 boys from 10 kids to prepare tea.

      {{{5/9}}}  is the probability to select one of remaining 5 boys from remaining 9 kids to cook.

      {{{C[4]^2/C[8]^2}}} = {{{6/28}}}  is the probability to select one of 6 possible pairs (girl,girl) 
                                                         from 28 possible pairs (kid,kid) to process table.
&lt;/pre&gt;

Solved in two ways, for your better understanding.



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A TWIN problem was solved at this forum several days ago under this link


https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1207851.html



</PRE>