Answer to Question #107435 in Statistics and Probability for Tarin

Question

The records of an Italian shoe manufacturer show that 20% of the shoes are
defective. Assume that selection of shoes follow a Bernoulli trial.
a. If a manufacturer inspects 15 shoes randomly, what is the probability that at
most two shoes are defective?
b. If a manufacturer wants to find 6 non defective shoes, what is the probability
that 12 shoes have to be inspected?
c. Find the expected number of inspected shoes to find 6 non defective shoes.

Answer 1

$a) P\{\xi\leq2\}=\sum_{s=0}^{2}C_{15}^{s}(0.2)^s(0.8)^{15-s}=0.398$ (Bernoulli's formula where $n=15, k=2, p=0.2$).

$b) P\{\eta=6\}=C_{12}^{6}(0.8)^6(0.2)^{6}=0.0155$ (Bernoulli's formula where $n=12, k=6, p=0.8$).

$c)$ $M\zeta-?$

$M\zeta=np=n(0.8)=6\\ n=7.5\\ (\text{if }\zeta \in B(n,p)\text{ then } M\zeta=np)$