ICs are packaged in boxes of 10. The probability of an IC being faulty is 2%. What is the probability of a box containing 2 faulty ics? Solution Given X~Binomial(n=10, p=0.02) P(X=x)=xC10*(0.02^x)*(0.98^(10-x)) So the probability is P(X=2)=2C10*(0.02^2)*(0.98^(10-2)) =(10*9/2)*(0.02^2)*(0.98^(10-2)) =0.01531373.