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tomo

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?

  • one year ago
  • one year ago

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  1. tomo
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    give me one second to show my work

    • one year ago
  2. tomo
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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) ??

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

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

    • one year ago
  5. satellite73
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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 \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

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

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

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

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

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

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

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

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

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

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

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

    • one year ago
  17. satellite73
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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

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

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

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

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

    • one year ago
  22. satellite73
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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

    • one year ago
  23. tomo
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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.

    • one year ago
  24. satellite73
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    i used \cap

    • one year ago
  25. tomo
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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 \[P(A|B) = P(A \cap B)/P(B)\]

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

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

    • one year ago
  28. satellite73
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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

    • one year ago
  29. satellite73
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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 \(P(A\cap B)=P(B)\times P(A|B)\)

    • one year ago
  30. satellite73
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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 \[P(A\cap B)=P(B)P(A|B)=.25\times .2\]

    • one year ago
  31. satellite73
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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

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

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

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

    • one year ago
  35. satellite73
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    ok

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

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

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

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

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

    • one year ago
  41. satellite73
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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

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

    • one year ago
  43. tomo
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    ok

    • one year ago
  44. satellite73
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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

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

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

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

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

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

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

    • one year ago
  51. satellite73
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    yw

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

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

    • one year ago
  54. satellite73
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    yw

    • one year ago
  55. tomo
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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?

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

    • one year ago
  57. tomo
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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?"

    • one year ago
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