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Thread: Problem with basic probability exercise. Any help will be very appreciated !

  1. #1
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    Problem with basic probability exercise. Any help will be very appreciated !

    Hi !
    I'm very unsure of my answers to this problem :

    A company produces boxes. 1% of the boxes have a defect. A control of the boxes detects and eliminates 95% of the defective boxes but eliminates by mistake 2% of the boxes produced.

    1) What's the probability not to eliminate a box, knowing it is not defective ?
    2) What's the probability the control doesn't work, i.e. that a non defective box is eliminated or a defective box is accepted ?
    3) What's the probability an accepted box is defective ?

    P(D) = 0,01
    P(E\D) = 0,95
    P(E n nonD) = 0,02


    1) P(nonE/nonD)=1-P(E/nonD)=1-0,02 = 0,98

    2) Pcontrol-failure = P(E n nonD)+P(nonE n D) - P(E n nonD) n (nonE n D) = p(D)*p(E/D) + p(nonD)*p(E/nonD) - P(E n nonD) n (nonE n D)
    Here I am stuck. I calculated that p(D)*p(E/D) + p(nonD)*p(E/nonD) = 0,0203. But for me the part I put in bold are exclusive events. However, I must be wrong because I was given 0,0195 as an answer.
    Are those events independant or exclusive ? I can't wrap my mind around this even though it's so basic I'm sorry...

    3) P(D/nonE) = P(nonE/D).P(D)/P(nonE)=(1- P(E/D)).P(D) / P(nonE)
    With P(nonE)= P(nonE/D).P(D) + P(nonE/nonD).P(nonD) = (1-P(E/D)).P(D) + P(nonE/nonD).P(nonD)


    Thank you for your help !!

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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    I'll just show you my solution and you can compare the two.

    The problem statement is a bit ambiguous. Does the control process remove 2% of non-defective boxes, or 2% of all boxes? I'm going to assume it's the first.

    we're told

    $P[D] = 0.01$ so $P[!D]=0.99$

    $P[E | D] = 0.95$

    $P[E | !D] = 0.02$

    1) you did correctly $P[!E|!D] = 1-P[E|!D] = 1-0.02 = 0.98$

    2) we want

    $P[(E \wedge !D) \vee (!E \wedge D)]$

    These are pretty clearly disjoint events so

    $P[(E \wedge !D) \vee (!E \wedge D)] = P[E | !D)]P[!D] + P[!E | D]P[D] = (0.02)(0.99)+(0.05)(0.01) = 0.0203$

    Which is what you got.

    3) we want

    $P[D|!E] = \dfrac{P[!E|D]P[D]}{P[!E]} = \dfrac{(0.05)(0.01)}{(0.05)(0.01)+(0.98)(0.99)} =5.15092\times 10^{-4}$

    I don't know what to tell you other than either we're interpreting the problem statement incorrectly or that the answer you were given is wrong.
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    Thank you for your answer !

    The assistant of my teacher replied to me and interpreted it as 2% of all boxes, so :
    P(E n nonD) = 0,02

    However I also find this very ambiguous... I translated from french, however this was just as ambiguous originally, and it was an exam exercise ahah.
    So I guess the answers are wrong in this case and the whole exercise has to be done differently.
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    Quote Originally Posted by Jo37 View Post
    Thank you for your answer !

    The assistant of my teacher replied to me and interpreted it as 2% of all boxes, so :
    P(E n nonD) = 0,02

    However I also find this very ambiguous... I translated from french, however this was just as ambiguous originally, and it was an exam exercise ahah.
    So I guess the answers are wrong in this case and the whole exercise has to be done differently.
    ok let's take another look then

    so 2% of all boxes are eliminated

    $P[E] = P[E|D]P[D] + P[E|!D]P[!D] = (0.95)(0.01) + P[E|!D](0.99) = 0.02$

    $P[E |!D] = \dfrac{0.02 - 0.0095}{0.99} = 0.0106061$

    substituting this new value for $P[E|!D]$ we get

    $P[(E\wedge !D) \vee (!E \wedge D)] = (0.0106061)(0.99)+(0.05)(0.01) = 0.011$

    still not matching the answer given.
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    I think there must be a mistake in the correction I was given.
    Thank you for your help and confirming the reasoning for question 2) in the first interpretation of the exercise !
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    Quote Originally Posted by Jo37 View Post
    I think there must be a mistake in the correction I was given.
    Thank you for your help and confirming the reasoning for question 2) in the first interpretation of the exercise !
    I am really bothered by this problem.
    Please check the translation of “A control of the boxes detects and eliminates 95% of the defective boxes”

    Have you studied the concept of false positives?
    Maybe if you posted the French, someone here can read it.
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    The original problem statement in french was :

    Une société produit des boîtes. 1% des boîtes produites sont défectueuses. Un contrôle de qualité automatisé permet de détecter et d'éliminer 95% des boîtes produites défectueuses mais élimine à tort 2% des boîtes produites.

    However the answer of 0,0195 was given to me by a member on another math forum. I then couldn't get this same result (and it is wrong) and noone ever replied back on the thread.
    My teacher confirmed this had to be interpreted as P(E n nonD) = 0,02.

    Yes, false positives would be non-defective boxes that the control considers defective by mistake ?
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    Quote Originally Posted by Jo37 View Post
    My teacher confirmed this had to be interpreted as P(E n nonD) = 0,02.
    so the first way we did it is correct? The answer of 0.0203?
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    The second way was the correct one !
    The ambiguity comes from "élimine à tort 2% des boîtes produites". It literally means word-by-word "eliminates wrongly 2% of produced boxes".
    I understood this as : 2% of good boxes are eliminated my mistake, since only good boxes can be eliminated by mistake, not bad ones which can only be accepted by mistake. So p(E |nonD).
    However it had to be read as out of all the boxes, 2% are eliminated by mistake.

    Wait, honestly my written explanation doesn't seem clear at all. ahah.
    I will send a final email to my teacher for clarification because this might cause me trouble for my exam.
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    Quote Originally Posted by Jo37 View Post
    The second way was the correct one !
    Ok, good.
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    Quote Originally Posted by Jo37 View Post
    A company produces boxes. 1% of the boxes have a defect. A control of the boxes detects and eliminates 95% of the defective boxes but eliminates by mistake 2% of the boxes produced.
    1) What's the probability not to eliminate a box, knowing it is not defective ?
    2) What's the probability the control doesn't work, i.e. that a non defective box is eliminated or a defective box is accepted ?
    3) What's the probability an accepted box is defective?

    Quote Originally Posted by Jo37 View Post
    The second way was the correct one !
    The ambiguity comes from "élimine à tort 2% des boîtes produites". It literally means word-by-word "eliminates wrongly 2% of produced boxes". However it had to be read as out of all the boxes, 2% are eliminated by mistake.
    My teacher confirmed this had to be interpreted as P(E n nonD) = 0,02.
    I absolutely agree that it is $\mathscr{P}(E\cap\neg D)=0.02.$
    So #1) becomes $\mathscr{P}(E|\neg D)=\dfrac{\mathscr{P}(E\cap\neg D)}{\mathscr{P}(\neg D)}=?$

    So #2) becomes $\mathscr{P}(E|\neg D)+\mathscr{P}(\neg E| D)=\dfrac{\mathscr{P}(E\cap\neg D)}{\mathscr{P}(\neg D)}+ \dfrac{\mathscr{P}(\neg E\cap D)}{\mathscr{P}(D)}=?$

    #3) Reads: "What is the probability that an accepted box is actually defective?"
    $\mathscr{P}(D|\neg E)=\dfrac{\mathscr{P}(D\cap\neg E)}{\mathscr{P}(E)}=~?$
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    Re: Problem with basic probability exercise. Any help will be very appreciated !

    1) 0.02 / 0.99

    2) 0.011 as romsek has calculated above.

    3) P(D/nonE) = = [P(nonE/D).P(D)] /P(nonE)

    I believe this is all correct now ! Thank you!

    PS : I posted a few threads at once trying to solve past exam subjects. I hope it is not contrary to the rules of the forum, I don't intend to flood.
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