Question 1205510
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A box contains 10 chocolates, of which 4 are dark chocolates and 6 are milk chocolates. 
If two chocolates are randomly selected without replacement, what is the probability 
of getting one dark chocolate and one milk chocolate in any order?
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<pre>
The number of all possible different pairs without looking the order 
is the number of combinations of 10 items taken 2 at a time 

    total = {{{C[10]^2}}}= {{{(10*9)/2}}} = 5*9 = 45.


The number of pairs (dark,milk), in this order, is 4*6 = 24;
the number of pairs (milk,dark), in this order, is 6*4 = 24;
but since the order does not matter, we DO NOT ADD 24 + 24 - we simply take/keep the value of 24
for all possible different pairs (dark,milk) or (milk,dark) of favorable pairs

    favorable = 6*4 = 24.


The probability under the problem's question is

    P = {{{favorable/total}}} = {{{24/45}}} = {{{8/15}}}.    <U>ANSWER</U>
</pre>

Solved.