I'm working on the problems related to Bayes' theorem - which did not have answers provided. I don't have confidence with my answer and ask if I can get some help/correction.
- You run the appliance department at your local ACME Superstore. Your toaster inventory is evenly split between those produced by the ACME corporation and General Products. ACME-branded toasters have a history of failing 9% of the time before the end of the warranty where as the General Products’ toasters fail 3% of the time before the end of the warranty.
a. A customer walks in with a broken toaster that they bought at your store, what is the probability that their toaster was made by ACME?
What I got:
P(ACME|Broken) = P(ACME&Broken) / P(Broken)
P(Broken) = 0.09( + 0.03 = 0.12
P(ACME&Broken) = (0.09/0.12) * 0.12 = 0.09
P(ACME|Broken) = 0.09 / 0.12 = 0.75
b. You make changes to your inventory such that you now have 2 ACME toasters for every 3 toasters made by General Products. Using the same scenario as question 1, what is the probability that the toaster brought in by the customer was made by ACME?
What I got:
P(ACME|Broken) = P(ACME&Broken) / P(Broken)
same equation but now P(ACME) = 0.4
P(ACME|Broken) = (0.4 * 0.12) / 0.12 = 0.4
For part $a$,
\begin{align}P(\text{broken})&=P(\text{broken}|\text{ACME})P(\text{ACME})+P(\text{broken}|\text{general})P(\text{general}) \\&=(0.09)(0.5)+(0.03)(0.5)\end{align}
Also, \begin{align}P(\text{ACME and broken)}&=P(\text{broken}|\text{ACME})P(\text{ACME}) \\ &=(0.09)(0.5)\end{align}
Try to make the corresponding correction for part b as well.
Edit:
For a, $$\frac{(0.09)(0.5)}{(0.09)(0.5)+(0.03)(0.5)}=\frac34$$
For b,
$$\frac{(0.09)(0.4)}{(0.09)(0.4)+(0.03)(0.6)}=\frac23$$