Question

A company that produces car breaks has a 15% defective rate. Suppose we are interested in the number of defectives in a random sample of size 8 of their products. (If we think of getting a defective break as a “success” then this is an example of a binomial experiment!) What is the probability that the number of defective breaks in the random sample is 2 or less?

Hint: You will need to use p(x) = n x p x (1 − p) n−x .

Answer 1

We have given,

$X\sim Binomial(n=8,p=0.15)$

$P[X\leq 2]$

=P[X=0]+ P[X=1]+P[X=2]

=0.2725+0.3847+0.2376

=0.8948

Where,

$P[X=0]=C(8,0)*(0.15)^{0}*(1-0.15)^{8}=0.2725$

$P[X=1]=C(8,1)*(0.15)^{1}*(1-0.15)^{7}=0.3847$

$P[X=2]=C(8,2)*(0.15)^{2}*(1-0.15)^{6}=0.2376$