Question

the probability of observing defective item in a lot is 0.01.suppose 25 items are inspected from the lot,what is the probability that not more than 3 defective are found, exactly 4 defective are found.(max 1024)

Answer 1

Let $X=$ the number of defective items: $X\sim Bin (n, p).$

Given $p=0.01, q=1-p=0.99, n=25.$

$$P(X\le3)=P(X=0)+P(X=1)$$

$$+P(X=2)+P(X=3)$$

$$=\dbinom{25}{0}(0.01)^0(0.99)^{25-0}+\dbinom{25}{1}(0.01)^1(0.99)^{25-1}$$

$$+\dbinom{25}{2}(0.01)^2(0.99)^{25-2}+\dbinom{25}{3}(0.01)^3(0.99)^{25-3}$$

$$=0.99989307347$$

$$P(X=4)=\dbinom{25}{4}(0.01)^4(0.99)^{25-4}$$

$$=0.00010243058$$