Question 1153867
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There are two ways to solve this problem.


One way is to use the formula  P = {{{C[6]^3/C[10]^3}}}.      (1)

The denominator of this formula,  {{{C[10]^3}}},  is the number of combinations of 10 items taken 3 at a time.

In other words, it is the number of all triples that can be formed from 10 pieces of fruits.

The numerator,  {{{C[6]^3}}},  is the number of all triples of pears that can be formed of 6 pears.


{{{C[10]^3}}} = {{{(10*9*8)/(1*2*3)}}} = 120.

{{{C[6]^3}}} = {{{(6*5*4)/(1*2*3)}}} = 20.


Therefore, the probability under the question is  P = {{{20/120}}} = {{{1/6}}}.



Another way is to write


    P = {{{(6/10)*(5/9)*(4/8)}}} = simplifying = {{{(2/10)*(5/3)*(1/2)}}} = {{{(1/3)*(1/2)}}} = {{{1/6}}},    (2)


which leads to the same answer.


In formula (2),  first factor  {{{6/10}}}  is the probability to get one pear from 10 fruits; 

the factor  {{{5/9}}}  is the probability to get second pear from the remaining 9 fruits; 

and the last factor  {{{4/8}}}  is the probability to get third pear from the remaining 8 fruits;.
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So, the problem can be solved by any of these two ways, and you should know both.