Conditional probability that a randomly chosen detail was made by Y, using Bayes's theorem

Question

I found the below question on the internet while working through a conditional probability questionnaire.

An automobile plant contracted to buy shock absorbers from two suppliers X and Y. X supplies 60% and Y supplies 40% of the shock absorbers. All shock absorbers are subject to quality tests and the ones that pass the tests are considered reliable. It was found that of X's supplies 96% were reliable and of Y's supplies 72% are reliable.

The Probability that a randomly chosen shock absorber, which is found to be reliable, is made by Y is:

(a) 0.288 (b) 0.334 (c) 0.667 (d) 0.720

I thought I could solve this question using the Bayes theorem. But as it turns out the answer I got by applying Bayes theorem was not right as I don't find it on the options. Could someone please help me find the right solutions as the website has only provided the questions and not the solutions.

Answer 1

Yes, Bayes's rule should work.

$$P(Y \mid \text{reliable}) = \frac{P(\text{reliable} \mid Y) P(Y)}{P(\text{reliable})} = \frac{P(\text{reliable} \mid Y) P(Y)}{P(\text{reliable} \mid X) P(X)+P(\text{reliable} \mid Y) P(Y)}$$

Plugging in the given values should give you (b).

Answer 2

Let $E$ be the event that a shock absorber is reliable. Then $$ Pr(Y|E)=\frac{Pr(Y\&E)}{Pr(E)}=\frac{Pr(E|Y)Pr(Y)}{Pr(E|Y)Pr(Y)+Pr(E|X)Pr(X)}=\frac{0.72*0.4}{0.72*0.4+0.96*0.6}. $$ The calculator says this equals $0.333333$ so choice (B)?

Answer 3

$$P\left[Y\mid R\right]=\frac{P\left[R\mid Y\right]P\left[Y\right]}{P\left[R\mid X\right]P\left[X\right]+P\left[R\mid Y\right]P\left[Y\right]}=\frac{0.72\times0.4}{0.96\times0.6+0.72\times0.4}=\frac{1}{3}\sim0.334$$

Here $R$ stands for reliable.