Conditional Probability problem

[lina29]

Homework Statement

A manufacturer of scientific workstations produces its new model at sites A, B, and C; 20% at A, 35% at B, and the remaining 45% at C. The probability of shipping a defective model is 0.01 if shipped from site A, 0.06 if from site B, and 0.03 if from site C.

A- What is the probability that a randomly selected customer receives a defective model?
B- If you receive a defective workstation, what is the probability that it was manufactured at site B?

Homework Equations

The Attempt at a Solution

For A I got .0365 which was correct but I'm stuck on part B. My assumption was that I had to find P(B|DB) where DB is being from site B and defective so I would use the equation
P(B^DB)/P(DB) I just don't know how I'm supposed to find P(B|DB) when I don't know what P(B^DB) is

 

[vela]

Homework Statement

A manufacturer of scientific workstations produces its new model at sites A, B, and C; 20% at A, 35% at B, and the remaining 45% at C. The probability of shipping a defective model is 0.01 if shipped from site A, 0.06 if from site B, and 0.03 if from site C.

A- What is the probability that a randomly selected customer receives a defective model?
B- If you receive a defective workstation, what is the probability that it was manufactured at site B?

Homework Equations

The Attempt at a Solution

For A I got .0365 which was correct but I'm stuck on part B. My assumption was that I had to find P(B|DB) where DB is being from site B and defective so I would use the equation
P(B^DB)/P(DB) I just don't know how I'm supposed to find P(B|DB) when I don't know what P(B^DB) is

You should just be calculating P(B|defective). The condition shouldn't specify where it came from. Think about it. If it's given that the workstation is defective and from site B, the probability it came from B is 1.

 

[lina29]

I'm confused are you saying I should be calculating
P(B|defective)= P(B^D)/P(D)=(.0365*.35)/(.0365)=.35 (which was counted wrong)
OR
that the probability is 1 which I don't get since the condition does specify that probability and there's not 100% chance it came from B since A & C have defective models also

 

[lina29]

Never mind I figured it out. Thanks!

 

[Ray Vickson]

I'm confused are you saying I should be calculating
P(B|defective)= P(B^D)/P(D)=(.0365*.35)/(.0365)=.35 (which was counted wrong)
OR
that the probability is 1 which I don't get since the condition does specify that probability and there's not 100% chance it came from B since A & C have defective models also

$$P(B \cap D) = P(D \cap B) = P(D|B) P(B).$$

RGV