Question 1194843
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A box contains 12 light bulbs of which 5 are defective. 
All the bulbs look alike and have equal probability of being chosen. 
Three bulbs are picked up at random. What is the probability that at least 2 are defective?
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The number of selecting 3 light bulbs from 12 is  {{{C[12]^3}}} = {{{(12*11*10)/(1*2*3)}}} = 220.


It goes to the denominator of the probability as the size of the space events.


The number of selecting 2 or more defective choosing 3 bulbs from 12 is

    {{{C[5]^2*C[12-5]^1}}} + {{{C[5]^3}}} = 10*7 + 10 = 80.


This number goes to the numerator of the probability as the size of favorable events.


So, the probability is  P = {{{favorable_triples/total_triples}}} = {{{80/220}}} = {{{4/11}}}.    <A>ANSWER</U>
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Solved.