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  1. #1
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    Question participent question

    Dear Sir,
    I really need help in the below question.
    Thanks
    Kingman

    In a test, questions are picked randomly. The probability that a participant gets an easy question is .6 while the probability of getting a difficult question is .4.

    The probability of any participant giving the correct answer to an easy question is .8 while the probability
    of giving the correct answer to a difficult question is .3.

    Find the probability that the participant give the correct answer twice in a row .

    I have 2 approaches to the question but wonder how to prove that the solution are the same using Venn diagram. Let

    E: an event that participant picked an easy question.
    D: an event that participant picked a difficult question.
    C: an event that the participant give the correct answer,
    W: an event that the participant give the wrong answer.

    Solution 1:
    Probability = (.6*.8+.4*.3)2 =.36 (using product rule)
    Solution 2:
    Using Tree diagram,
    Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36
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  2. #2
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    Quote Originally Posted by kingman View Post
    Dear Sir,
    I really need help in the below question.
    Thanks
    Kingman

    In a test, questions are picked randomly. The probability that a participant gets an easy question is .6 while the probability of getting a difficult question is .4.

    The probability of any participant giving the correct answer to an easy question is .8 while the probability
    of giving the correct answer to a difficult question is .3.

    Find the probability that the participant give the correct answer twice in a row .

    I have 2 approaches to the question but wonder how to prove that the solution are the same using Venn diagram. Let

    E: an event that participant picked an easy question.
    D: an event that participant picked a difficult question.
    C: an event that the participant give the correct answer,
    W: an event that the participant give the wrong answer.

    Solution 1:
    Probability = (.6*.8+.4*.3)2 =.36 (using product rule)
    Solution 2:
    Using Tree diagram,
    Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36
    Why do you need to use a Ven Diagram for this?

    I would use a unit square divided into parts in the proportions given in the question, then the required probability will be the area corresponding to a correct answer. This is virtually imposible to describe without a diagram.

    CB
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  3. #3
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    Question use set operations

    What I mean is to use set operation like union, intersection and P( A) ,P(A') P(c/E) set notations to represent the probabilities found in the two solutions and prove the probabilities of the events are equal.
    Thanks
    Kingman
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by kingman View Post
    What I mean is to use set operation like union, intersection and P( A) ,P(A') P(c/E) set notations to represent the probabilities found in the two solutions and prove the probabilities of the events are equal.
    Thanks
    Kingman
    Small note: If you have a keyboard like mine, you can find the key for | right above the Enter/Return key, as the shifted character above backslash (\).

    Using your letters, we have

    P(E) = 0.6

    P(D) = 0.4

    P(C|E) = 0.8

    P(C|D) = 0.3

    So what you're asking for is

    P(C) = P(C \cap (E \cup D))

    = P((C \cap E) \cup (C \cap D))

    = P(C \cap E) + P(C \cap D) - P((C \cap E) \cap (C \cap D))

    = P(C \cap E) + P(C \cap D)

    = P(E)\cdot P(C|E) + P(D)\cdot P(C|D)

    = (0.6)(0.8) + (0.4)(0.3)

    = \frac{3}{5}

    Then the probability of this happening two times in a row is

    P(X) = \left(\frac{3}{5}\right)^2 = \frac{9}{25} = 0.36
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  5. #5
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    Dear Undefined,
    Thanks very much for the answer .
    Can you show me how to prove the term after the minus sign of the below expression is zero as you have written in your answer:

    P(C)=

    and can you show me how to prove the second solution as given in the question:
    Using Tree diagram,
    Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 *(.6*.8+.4*.3) =.36


    * (P(E). P(C|E) +P(D).P(C|D))

    also I wonder how can write the union and set symbol
    in this message window?

    Thank you very much
    Kingman
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by kingman View Post
    Dear Undefined,
    Thanks very much for the answer .
    Can you show me how to prove the term after the minus sign of the below expression is zero as you have written in your answer:

    P(C)=
    E and D are mutually exclusive events. The probability of them happening at the same time is zero.

    Quote Originally Posted by kingman View Post
    and can you show me how to prove the second solution as given in the question:
    Using Tree diagram,
    Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 *(.6*.8+.4*.3) =.36


    * (P(E). P(C|E) +P(D).P(C|D))
    I'm not sure what you're asking. Didn't you have to draw the tree diagram in order to write down the expression? Or did you just copy the expression from somewhere? I originally assumed you had discovered the solution yourself, but I guess you got it from the back of the book or some such.

    Try drawing the tree diagram and post again if you have trouble.

    But one critique: I think it's overkill to draw a whole diagram to go along with the expression. Since the participant's answering the first question and second question are independent events, it's enough to find the probability of answering one question correctly, then square the result.

    Quote Originally Posted by kingman View Post
    also I wonder how can write the union and set symbol
    in this message window?

    Thank you very much
    Kingman
    See the LaTeX tutorial thread, which is within the LaTeX Help subforum. The LaTeX command for union is \cup and for intersection is \cap.

    There are also unicode symbols but I don't know the codes to enter them with a keyboard. They look like this and can be copy and pasted: ∪ ∩
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  7. #7
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    need help

    Dear Undefined,
    I have drawn the tree diagram and saved its image in gif. format and I need help to know how to
    send or paste the file in message window to you .
    Thanks
    Kingman
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by kingman View Post
    Dear Undefined,
    I have drawn the tree diagram and saved its image in gif. format and I need help to know how to
    send or paste the file in message window to you .
    Thanks
    Kingman
    When you are posting a reply, look down a little where it says "Additional Options" and then "Attach Files" where there is a button "Manage Attachments." Then you can choose "Browse" to find the file on your computer and then "Upload." Then it should appear on this forum!
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  9. #9
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    Thanks

    Dear Undefined,
    Thanks for the guidance and I think I have manage to attach the tree diagram into this message box.
    I would appreciate very much if you can show me how to write the required probability from the tree diagram and show that it is actually looks like


    (P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))

    Thanks
    Kingman
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  10. #10
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by kingman View Post
    Dear Undefined,
    Thanks for the guidance and I think I have manage to attach the tree diagram into this message box.
    I would appreciate very much if you can show me how to write the required probability from the tree diagram and show that it is actually looks like


    (P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))

    Thanks
    Kingman
    Okay, so your diagram accurately matches the expression given in the first post, that is,

    .6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36

    I've re-attached your image with a red rectangle on it; you can draw just what's in the red rectangle to be able to write

    P(X) = .6*.8+.4*.3

    where X is the event of answering a question correctly.

    Then, without drawing any diagrams, you can reason that since answering the first question correctly is independent from answering the second question correctly, we may simply write that the desired probability is

    P(X)*P(X)

    This is because the situation is just like rolling a die two times in a row, and asking what is the probability of rolling a 3 two times in a row. Rolling a three once has probability (1/6). Doing it twice in a row has probability (1/6)(1/6) = (1/36).

    Edit: Forgot to attach image. Also, some minor typos/etc.
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  11. #11
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    Thanks very much

    Dear Undefined.

    Thanks very much for the answer but I wonder whether you can show me
    how to prove that the Probability of the four events indicated as 4 loops in the attached diagram
    equals to
    (P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))

    What you have proven is for solution 1 but what is needed is solution 2 equivalent.
    Thanks
    Kingman
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  12. #12
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by kingman View Post
    Dear Undefined.

    Thanks very much for the answer but I wonder whether you can show me
    how to prove that the Probability of the four events indicated as 4 loops in the attached diagram
    equals to
    (P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))

    What you have proven is for solution 1 but what is needed is solution 2 equivalent.
    Thanks
    Kingman
    Well what I was saying is, in order to get that expression, don't use the diagram as you attached it, but only use the part enclosed in red as I attached it (above). There are two leaves marked C (a leaf is an end node). So one leaf corresponds to P(E). P(C|E) and the other leaf corresponds to P(D).P(C|D). You add those two. And then you square the result. Try plugging in the numbers.

    Quote Originally Posted by undefined View Post
    P(E) = 0.6

    P(D) = 0.4

    P(C|E) = 0.8

    P(C|D) = 0.3
    You will see that it works out as expected.
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  13. #13
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    Thanks

    Dear Undefined.

    thanks for the answer and I would be grateful if you can show me how to find the
    probability of the 4 events :
    P( (ECEC) + P( ECDC) + P(DCEC) + P( DCDC) which should be equal to

    (P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))


    Thanks
    Kingman
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  14. #14
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by kingman View Post
    Dear Undefined.

    thanks for the answer and I would be grateful if you can show me how to find the
    probability of the 4 events :
    P( (ECEC) + P( ECDC) + P(DCEC) + P( DCDC) which should be equal to

    (P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))


    Thanks
    Kingman
    Your notation is non-standard. P(ECEC) isn't something people tend to write, but anyway I know what you mean.

    You wrote down numbers on the edges. You multiply these numbers. So,

    "P(ECEC) + P(ECDC) + P(DCEC) + P(DCDC)" = 0.6*0.8*0.6*0.8 + 0.6*0.8*0.4*0.3 + 0.4*0.3*0.6*0.8 + 0.4*0.3*0.4*0.3

    Get out your calculator and you will find this equals 0.36.

    You know this is equal to

    (P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))

    because when you plug in the numbers and get out your calculator and work it out, you also get 0.36!
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  15. #15
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    thanks

    Dear Undefined

    Numerically I can verify that the answer is correct but I do not how to represent the product of number in Probability of event form ;
    eg

    "P(ECEC) + P(ECDC) + P(DCEC) + P(DCDC)" = 0.6*0.8*0.6*0.8 + 0.6*0.8*0.4*0.3 + 0.4*0.3*0.6*0.8 + 0.4*0.3*0.4*0.3

    in the first term : 0.6*0.8*0.6*0.8
    For P(ECEC)
    P(E)=.6
    P(C|E)=.8
    then how to write P( ..of what....)=.6
    and P( of what.....)=.8

    I do not know whether writing P(C|E and C)=.6 is correct
    and whether P(C|E and C and E)=.8 is correct

    If it is correct the how to prove that the sum equals to
    (P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D)) algebraically.
    thanks
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