Question -
Answer -
Let there be x numberof defective items in a sample of ten items drawn successively.
Now, as we can seethat the drawing of the items is done with replacement. Thus, the trials areBernoulli trials.
Now, probability ofgetting a defective item, p = 5/100 = 1/20
Thus, q = 1 – 1/20 =19/20
∴ We can say thatx has a binomial distribution, where n = 10 and p = 1/20
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2 …n
Probability of gettingnot more than one defective item = P(X ≤1)
= P(X = 0) + P(X = 1)
= 10C0 (19/20)10(1/20)0 +10C1 (19/20)9(1/20)1
