2% of a product are defective. If a lot of 100 items are ordered what is the probability that there are no defective items? What is the probability that there are at least two defective items?

Guest Feb 18, 2020

#1**+1 **

This is a binomial distribution problem.

Use the formula: _{n}C_{r}·p^{r}·q^{n-r}

where: n = number to choose from r = number chosen p = probability of "success"

q = probability of "failure" = 1 - r

For the first problem: n = 100 r = 0 p = 0.02 q = 0.98

---> Answer = _{100}C_{0}·(0.02)^{0}·(0.98)^{100}

For the second problem: "at least two" means "two or more" and then you would need to use the above formula for 2, for 3, for 4, ... for 100 which would be horribly long. Instead, find the probability for "fewer than two" [that is, zero or one] and subtract this from 1 [100%].

---> Answer = 1 - _{100}C_{0}·(0.02)^{0}·(0.98)^{100} - _{100}C_{1}·(0.02)^{1}·(0.98)^{99}

geno3141 Feb 18, 2020

#1**+1 **

Best Answer

This is a binomial distribution problem.

Use the formula: _{n}C_{r}·p^{r}·q^{n-r}

where: n = number to choose from r = number chosen p = probability of "success"

q = probability of "failure" = 1 - r

For the first problem: n = 100 r = 0 p = 0.02 q = 0.98

---> Answer = _{100}C_{0}·(0.02)^{0}·(0.98)^{100}

For the second problem: "at least two" means "two or more" and then you would need to use the above formula for 2, for 3, for 4, ... for 100 which would be horribly long. Instead, find the probability for "fewer than two" [that is, zero or one] and subtract this from 1 [100%].

---> Answer = 1 - _{100}C_{0}·(0.02)^{0}·(0.98)^{100} - _{100}C_{1}·(0.02)^{1}·(0.98)^{99}

geno3141 Feb 18, 2020