SOLUTION: There are 10 pcs of fruit, 6 were pears. If worker randomly selects 3 pcs for a basket, what is the probability that all were pears?

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Question 1153867: There are 10 pcs of fruit, 6 were pears. If worker randomly selects 3 pcs for a basket, what is the probability that all were pears?
Answer by ikleyn(52814) About Me  (Show Source):
You can put this solution on YOUR website!
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There are two ways to solve this problem.


One way is to use the formula  P = C%5B6%5D%5E3%2FC%5B10%5D%5E3.      (1)

The denominator of this formula,  C%5B10%5D%5E3,  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%5B6%5D%5E3,  is the number of all triples of pears that can be formed of 6 pears.


C%5B10%5D%5E3 = %2810%2A9%2A8%29%2F%281%2A2%2A3%29 = 120.

C%5B6%5D%5E3 = %286%2A5%2A4%29%2F%281%2A2%2A3%29 = 20.


Therefore, the probability under the question is  P = 20%2F120 = 1%2F6.



Another way is to write


    P = %286%2F10%29%2A%285%2F9%29%2A%284%2F8%29 = simplifying = %282%2F10%29%2A%285%2F3%29%2A%281%2F2%29 = %281%2F3%29%2A%281%2F2%29 = 1%2F6,    (2)


which leads to the same answer.


In formula (2),  first factor  6%2F10  is the probability to get one pear from 10 fruits; 

the factor  5%2F9  is the probability to get second pear from the remaining 9 fruits; 

and the last factor  4%2F8  is the probability to get third pear from the remaining 8 fruits;.


So, the problem can be solved by any of these two ways, and you should know both.