# Thread: Help on Baye's theorem problem

1. ## Help on Baye's theorem problem

A company request materials from three stores X, Y and Z. Store X supplies 45% of the material, store Y supplies 30% of the materials while store Z supplies the remainder. Investigation has shown that 12% of the materials supplied from store X is defective, 15% of the materials supplied by store Y is defective while only 6% of the materials supplied by stores Z is defective.

If an item is returned after being purchased due to a defect, use Baye’s theorem to calculate the probability that the defective materials was supplied by store
i. X
ii.Z

2. ## Re: Help on Baye's theorem problem

Originally Posted by benedec
A company request materials from three stores X, Y and Z. Store X supplies 45% of the material, store Y supplies 30% of the materials while store Z supplies the remainder. Investigation has shown that 12% of the materials supplied from store X is defective, 15% of the materials supplied by store Y is defective while only 6% of the materials supplied by stores Z is defective. If an item is returned after being purchased due to a defect, use Baye’s theorem to calculate the probability that the defective materials was supplied by store
i. X ii.Z
$P(D)=P(D\cap X)+P(D\cap Y)+P(D\cap Z)$
$P(D)=P(D|X)P(X)+P(D|Y)P(Y)+P(D|Z)P(Z)$.

Now you want to find $P(X|D)=\frac{P(D\cap X)}{P(D)}$

3. ## Re: Help on Baye's theorem problem

Thank you Plato. In a layman terms, can you elaborate more

4. ## Re: Help on Baye's theorem problem

Originally Posted by benedec
Thank you Plato. In a layman terms, can you elaborate more
Are you studying probability or not?
If you are you have a textbook and a set of notes.
If you are not then there no point in posting a probability question.

5. ## Re: Help on Baye's theorem problem

Reasons I asked it seems the formula used is not same as mine, this is what I tried

P (defective | X) = P (defective | X) P(warehouse X)
P (defective | X) P(X)+ P (defective | Y) P(Y )+ P (defective | Z) P(Z)
= 1/3 x 0.12 = 0.33 x 0.12
1/3 x 0.12 + 1/3 x 0.15 + 1/3 x 0.06 0.33 x 0.12 + 0.33 x 0.15 + 0.33 x 0.66
= 0.0396 = 0.0396 = 0.129
0.0396 + 0.0495 + 0.2178 0.3069

6. ## Re: Help on Baye's theorem problem

Originally Posted by benedec
Reasons I asked it seems the formula used is not same as mine, this is what I tried

P (defective | X) = P (defective | X) P(warehouse X)
P (defective | X) P(X)+ P (defective | Y) P(Y )+ P (defective | Z) P(Z)
= 1/3 x 0.12 = 0.33 x 0.12
1/3 x 0.12 + 1/3 x 0.15 + 1/3 x 0.06 0.33 x 0.12 + 0.33 x 0.15 + 0.33 x 0.66
= 0.0396 = 0.0396 = 0.129
0.0396 + 0.0495 + 0.2178 0.3069

Now you want to find $P(X|D)=\frac{P(D| X)P(X)}{P(D|X)P(X)+P(D|Y)P(Y)+P(D|Z)P(Z)}$