Question

2. Suppose that a large lot of lambs contains 17% defectives. Find the probability that a random sample of 8 fuses contains: a) Exactly two defective, b) At most two defective, b) At least one defective.

Answer 1

Solution:

We are given that a large lot of lambs contains 17% defectives. Therefore, we have:

$p=0.17$

Let x be the number of defectives

We know that a random variable x follows a binomial distribution, with parameters $n=8,p=0.17$

a) Exactly two defectives.

Answer:  

$P(x=2)=\binom{8}{2}0.17^{2}(1-0.17)^{8-2}$

$=\frac{8!}{(8-2)!2!}0.17^{2}(1-0.17)^{8-2}$

$=28 \times 0.0289 \times 0.326940373$

$=0.2646$

Therefore, the probability of exactly two defectives is 0.2646.

b) At least one defective.

Answer:  

$P(x\geq1)=1-P(x<1)$

$=1-P(x=0)$

$=1- \binom{8}{0}0.17^{0}(1-0.17)^{8-0}$

$=1-0.2252$

$=0.7748$

Therefore, the probability of at least one defective is 0.7748.

c) At most two defective

Answer:

$P(x \leq 2)=P(x=0)&plus;P(x=1)&plus;P(x=2)$

$=\binom{8}{0}0.17^{0}(1-0.17)^{8-0}&plus;\binom{8}{1}0.17^{1}(1-0.17)^{8-1}&plus;\binom{8}{2}0.17^{2}(1-0.17)^{8-2}$

$=0.225229&plus;0.369050&plus;0.264560$

$=0.8588$

Therefore, the probability of at most two defectives is 0.8588