Results 1 to 3 of 3

Math Help - Probability problem

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    60

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Probability problem

    Quote Originally Posted by Schdero View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    60

    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Replies: 0
    Last Post: October 8th 2009, 08:45 AM
  3. Help with a Probability problem!
    Posted in the Statistics Forum
    Replies: 3
    Last Post: January 28th 2009, 11:59 AM
  4. probability problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 16th 2008, 12:45 PM
  5. probability problem
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 2nd 2008, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum