Question 1190737
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During the war one truck out of nine was destroyed due to bombardment. 
What was the probability that exactly three out of six trucks would arrive safely ?
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It is a typical binomial distribution problem.


The probability of the success (arrive safely) in each single trial is 1/9 = 0.111111.


The number of trials is  n= 6;  the number of successful trials is  k= 3.


The formula for the probability is


    P(n= 6; k= 3; p= 0.111111) = {{{C[6]^3*0.111111^3*(1-0.111111)^(6-3)}}} = {{{20*0.111111^3*0.888889^3}}} = 0.019268316,  or  0.01927  (rounded).    <U>ANSWER</U>
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Solved.