1. ## Probability problem

A machine produces a light bulb in two independent steps (A and B). A light bulb is to be considered flawed, if in one of the two steps an error occurs. It is known, that out of 200 produced light bulbs 16 are flawed. It is farther known, than in the first step the probability for an error to occur is 1/24 (P(A)=1/24). What is the probability of an error in step 2?

The official solution is P(AorB)=P(A)+P(B)-P(AandB) wheras P(AandB)= P(A)*P(B) as A,B are independent. Thus P(AorB)=P(A)+P(B)-P(A)*P(B). What they do now is they insert for P(A)=1/24 and for P(AorB)=16/200. My problem with that is, that for a light bulb to be considered flawed, it could also well be that in both steps an error occured, thus 16/200 is, so I think, not P(AorB) but rather P(AorB)+P(AandB). Maybe this doesnt matter, as the steps are independent though, im not sure. If you solve this equation for P(B) you get 0.03996.

My solution was another, as I looked at it from another angle: I thought the possibility of NOT making an error is 184/200 in general and 23/24 for step A. Thus the Possibility for NOT making an error in B must be 184/200=(23/24)*P(B'). If you solve that for P(B') you get P(B')= 0.96 and hence P(B)=1-P(B')=0.04.

What do you think of my soultion? And what about the problem, that the original solution the error probability = P(AorB) and not P(AorB) + P(AandB)?

Thank you so much

2. ## Re: Probability problem

Originally Posted by Schdero
A machine produces a light bulb in two independent steps (A and B). A light bulb is to be considered flawed, if in one of the two steps an error occurs. It is known, that out of 200 produced light bulbs 16 are flawed. It is farther known, than in the first step the probability for an error to occur is 1/24 (P(A)=1/24). What is the probability of an error in step 2?

The official solution is P(AorB)=P(A)+P(B)-P(AandB) wheras P(AandB)= P(A)*P(B) as A,B are independent. Thus P(AorB)=P(A)+P(B)-P(A)*P(B). What they do now is they insert for P(A)=1/24 and for P(AorB)=16/200. My problem with that is, that for a light bulb to be considered flawed, it could also well be that in both steps an error occured, thus 16/200 is, so I think, not P(AorB) but rather P(AorB)+P(AandB). Maybe this doesnt matter, as the steps are independent though, im not sure. If you solve this equation for P(B) you get 0.03996.

My solution was another, as I looked at it from another angle: I thought the possibility of NOT making an error is 184/200 in general and 23/24 for step A. Thus the Possibility for NOT making an error in B must be 184/200=(23/24)*P(B'). If you solve that for P(B') you get P(B')= 0.96 and hence P(B)=1-P(B')=0.04.

What do you think of my soultion? And what about the problem, that the original solution the error probability = P(AorB) and not P(AorB) + P(AandB)?

Thank you so much
You do know that the logical connective "or" means?

"A or B" means one or both of {A,B}

CB

3. ## Re: Probability problem

Actually I didn't, thank you. But is the difference between my solution and the original solution just due to rounding errors or is my solution wrong?

Thank you