Answer :

(i) The probability that the item is defective = = p


The probability that the bulb will not fuse = 1= = q


Using Bernoulli’s we have,


P(Success=x) = nCx.px.q(n-x)


x=0, 1, 2, ………n and q = (1-p), n =6


The probability that exactly 2 defective items are,


6C2.()2()4



(ii) The probability that the item is defective = = p


The probability that the bulb will not fuse = 1= = q


Using Bernoulli’s we have,


P(Success=x) = nCx.px.q(n-x)


x=0, 1, 2, ………n and q = (1-p), n =6


The probability that not more than 2 defective items are,


6C0.()0()6 + 6C1.()1()5 + 6C2.()2()4



(iii) The probability that the item is defective = = p


The probability that the bulb will not fuse = 1= = q


Using Bernoulli’s we have,


P(Success=x) = nCx.px.q(n-x)


x=0, 1, 2, ………n and q = (1-p), n =6


The probability of at least 3 defective items are,


P(3) + P(4) + P(5) + P(6)


6C3.()3()3 + 6C4.()4()2 + 6C5.()5()1 + 6C6.()6()0



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