1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditional Probability

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Question 1:
    A teacher gave his class two tests where every student passed at least one test. 72% of the class passed passed both tests and 80% of the class passed the second test.
    (i)what percentage of those who passed the second test also passed the first?
    (ii) what percentage of the class passed the first but failed the second test?

    Question 2
    A factory has three machines P,Q and R, producing large numbers of a certain item Of the total production, 40% is produced on P, 50% on Q and 10% on R. The records show that 1% of the items produced on P are defective, 2% of items produced on Q are defective and 6% of items of items produced on R are defective The occurrence of a defective item is independent of each machine and all other items.
    (i)calculate the probability the item chosen is defective.
    (ii) Given that the item chosen is defective, find the probability that it was produced on machine
    Q.
    2. Relevant equations

    P(A|B) = P(AnB)/P(B)


    3. The attempt at a solution

    1. (i)

    P(F|S) = .72/.8 = .9

    (ii)

    Since all students pass at least one test.

    P(F U S) = 1

    1= P(F) + P(S) - P(FnS).......... P(F) = 0.92 all people who passed the first test

    P(F) - P(FnS) = 0.2 This is the number who only passed the first test.

    According to the answers in my book I got the first part right but the second part wrong.

    Question 2
    (i)
    PD = .4(.01)
    QD = .5(.02)
    RD = .1(.06)

    Adding all three gives .02

    (ii)

    P(Q|D) = .5(.02)/.5 = .02

    According to my book both of these answers are wrong.Any help would be appreciated.
     
  2. jcsd
  3. Sep 19, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I'd say the book is wrong.
    Looks right to me.
    Think again about what you are dividing by here.
     
  4. Sep 19, 2013 #3
    Would it be P(Q|D) = .5(.02)/.02 = .5 ??
     
  5. Sep 20, 2013 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes. Do you see how that follows from the equation?
     
  6. Sep 20, 2013 #5
    Ya, I didnt divide by the probability of q being defective the first time. Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Conditional Probability
  1. Conditional probability (Replies: 10)

Loading...