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tomo

  • one year ago

A multiple-choice test has five possible answers for each questions. On any given question a student has probability .75 to know the right answer and probability .25 to have to guess. Since there are five possible answers, the probability he guesses correctly is .25. If a student answered the question correctly, what is the probability that he guessed?

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  1. tomo
    • one year ago
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    give me one second to show my work

  2. tomo
    • one year ago
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    P(knows correct) = .75 P(guess) = .25 Is the probability he guesses correctly .25 or is it 1/5 = .2? I believe that i am looking for P(guess given that it is correct) ??

  3. satellite73
    • one year ago
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    you are looking for a conditional probability

  4. satellite73
    • one year ago
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    and i believe you need baye's formula for this one

  5. satellite73
    • one year ago
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    if you put \(A\) as the probability he gets the answer right, and \(B\) as the probability he guessed, you are looking for \[P(A|B)\] which you compute via \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

  6. satellite73
    • one year ago
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    ok that was wrong, you are looking for \(P(B|A)\)

  7. tomo
    • one year ago
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    yes, P(guess given that he answers correctly)

  8. satellite73
    • one year ago
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    using the sets i wrote above, you compute \[P(B|A)=\frac{P(A\cap B)}{P(A)}\]

  9. tomo
    • one year ago
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    so, can we say that P(guess AND correct) = P(guess) * P(correct)

  10. satellite73
    • one year ago
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    yes, that is your numerator

  11. satellite73
    • one year ago
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    oh wait no it isn't sorry

  12. tomo
    • one year ago
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    that is assuming the two are independent right

  13. satellite73
    • one year ago
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    because they are not independent

  14. tomo
    • one year ago
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    so do i have to use the extended bayes formula

  15. satellite73
    • one year ago
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    you compute \[P(A\cap B)=P(B)\times P(A|B)\] yes, it is baye's formula you need

  16. satellite73
    • one year ago
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    both of those are easy to compute, \(P(B)=.25\) and \(P(A|B)=.2\) i think

  17. satellite73
    • one year ago
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    this line from your question . Since there are five possible answers, the probability he guesses correctly is .25. is wrong i think

  18. satellite73
    • one year ago
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    if there are 5 possible answers, and he guesses, the probability he guesses right should be \(.2\) not \(.25\)

  19. tomo
    • one year ago
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    I'm confused. Is P(B AND A) the same as P(A AND B)?

  20. satellite73
    • one year ago
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    i think you should get for the numerator \(.25\times .2\) and for the denominator \[.25\times .2+.75\times 1\]

  21. satellite73
    • one year ago
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    yes, they are the same \(P(A\cap B)=P(B\cap A)\)

  22. satellite73
    • one year ago
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    you know how to compute \(P(A)\) the probability he guesses, and \(P(B|A)\) the probability he gets it right given that he guesses

  23. tomo
    • one year ago
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    where do i find the AND/intersection statement in the equation editor? i can't find it and it would help me explain my thinking.

  24. satellite73
    • one year ago
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    i used \cap

  25. tomo
    • one year ago
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    thanks, ok I'm confused because I thought that changing the order of B and A in bayes formula changed the B and A in the intersection. \[P(B|A) = P(B \cap A)/P(A) \] and \[P(A|B) = P(A \cap B)/P(B)\]

  26. satellite73
    • one year ago
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    let me get to a desktop, it is hard to write this on a lap top

  27. satellite73
    • one year ago
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    ok that's better

  28. satellite73
    • one year ago
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    both the formulas you wrote above are correct, and the numerators are both the same the point of baye's is that you can use the conditional probability you know (the easy one) to compute the conditional probability you want (the hard one) at the moment, we don't know \(P(A\cap B)\) so we have to use a conditional probability to find it

  29. satellite73
    • one year ago
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    if \(A\) is you get the right answer, and \(B\) is you guess, we can know both \(P(B)=.25\) because it is given to us, and we can compute \(P(A|B)\) by thinking so we can compute \(P(A\cap B)=P(B)\times P(A|B)\)

  30. satellite73
    • one year ago
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    \(P(A|B)\) means you guess, and you get the answer correct. since one out of the 5 answers is correct, if you guess you have a one in five (aka .2\) chance of getting it right, so \(P(A|B)=.2\) and therefore \[P(A\cap B)=P(B)P(A|B)=.25\times .2\]

  31. satellite73
    • one year ago
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    we cannot compute \(P(B|A)\) because it is not obvious, it is what we are looking for so we have to use the one above

  32. satellite73
    • one year ago
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    now we need \(P(A)\) the denominator.

  33. satellite73
    • one year ago
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    btw after we do this the hard way i will show you a very quick snappy way to do it

  34. tomo
    • one year ago
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    ok, give me one second to compute the answer.

  35. satellite73
    • one year ago
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    ok

  36. satellite73
    • one year ago
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    then we will do it in our heads with almost no computation at all

  37. tomo
    • one year ago
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    P(B|A) = .05/.75 = .0666

  38. satellite73
    • one year ago
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    that is not what i get

  39. satellite73
    • one year ago
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    your denominator is wrong it should be \(P(A)=.75+.25\times .2=.8\)

  40. tomo
    • one year ago
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    oh yeah, I forgot to add the probability of guessing correctly

  41. satellite73
    • one year ago
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    \(A\) is he gets it right, but he gets it right more than 75% of the time he gets it right if he knows it and also if he guesses correctly

  42. satellite73
    • one year ago
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    ok now lets do it the snap way

  43. tomo
    • one year ago
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    ok

  44. satellite73
    • one year ago
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    he answers 100 questions, and he gets 75 right because he knows them and of the remaining 25 he gets 5 right

  45. satellite73
    • one year ago
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    therefore he gets 80 right total, and of those 80 he guessed on 5 your answer is therefore \(\frac{5}{80}\)

  46. satellite73
    • one year ago
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    wasn't that easy?

  47. tomo
    • one year ago
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    interesting. I didn't think of that.

  48. satellite73
    • one year ago
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    baye's formula becomes much less mysterious when you use numbers and think

  49. tomo
    • one year ago
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    thank you very much for all the help. I really appreciate it. Sometimes this probability can be quite confusing.

  50. satellite73
    • one year ago
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    easier than saying \[P(B|A)=\frac{P(B)P(A|B)}{P(B)P(A|B)+P(B^c)P(A|B^c)}\]

  51. satellite73
    • one year ago
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    yw

  52. satellite73
    • one year ago
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    hint, on an exam if you get confused use numbers make them big

  53. tomo
    • one year ago
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    thanks for the tip.

  54. satellite73
    • one year ago
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    yw

  55. tomo
    • one year ago
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    A little way up, you said that \[P(A|B) = .2\] I was thinking about that and tried doing a probability tree. |dw:1358740038530:dw| why isn't it .25*.2. (.25 for guessing and .2 for guessing it correctly.)? Does \[P(A|B) \] mean probability of B given that we know A has occurred? Now that be occurs what is the probability of A?

  56. tomo
    • one year ago
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    be should be 'B'

  57. tomo
    • one year ago
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    right after P(A|B) i messed up. I mean "Does P(A|B) mean probability of A given that we know B has occurred? Now that B occurs what is the probability of A?"

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