A manufactured product is either produced intact or produced with a fault. The probability of a fault is 0.1. An inspection of 10 products, chosed at random is performed. If at least two of the ten products are found to have a fault, an alarm is raised.
i) Calculate the probability that an alarm is raised during an inspection.
My attempt for (i):
We have that $P(fault)=0.1$.
$P(alarm)=P(\text{ at least 2 faulty })=1-P(\text{ 9 or 10 not faulty })=1-(P(\text{ 9 not faulty })+P(\text{ 10 not faulty }))=1-(\frac{10!}{9!}(0.1)(0.9)^9+(0.9)^{10})=1-0.38-0.34=0.28$
(ii) Given an alarm is raised, what is the probability that there are exactly two faulty products in the inspected sample?
My attempt for (ii):
$P(\text{ exactly 2 faults } | \text{ alarm })=\frac{ P( \text{ exactly 2 faults } \cap \text { alarm })}{P(\text{ alarm })}=\frac{ P( \text{ exactly 2 faults } )}{P(\text{ alarm })}= \frac{\frac{10!}{8!2!}(0.1)^2(0.9)^8}{0.28}=\frac{0.19}{0.28}=0.67$
(iii) Calculate the probability that in 5 independent inspection (each of 10 products), there is at least one alarm raised.
My attempt:
$P(\text{ at least 1 alarm in 5 inspections })=1-P(\text{ no alarm in 5 inspections }) = 1-(1-P(\text{ alarm })^5=1-(1-0.28)^5=0.81$
Is all i've done correct?
Come on!!!
Your exercise is perfect....quite perfect :(
YOU CANNOT APPROX a probability as you did....
Which lead to
and
as you can see the numerical results are very different form yours.