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gorv

  • one year ago

To get an A grade on the test, you need a total score of more than 16 points. One of the students knows the correct answer to 6 of the 20 questions. The rest she guesses at random by tossing a coin (one toss per question, as in 4B). What is the chance that she gets an A grade on the test?

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  1. gorv
    • one year ago
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    if the coin lands Heads, she answers True, and if it lands Tails, she answers False

  2. Jack1
    • one year ago
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    one point per question, yeah?

  3. Jack1
    • one year ago
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    9%...?

  4. Jack1
    • one year ago
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    P of getting 16/20 or more : knows 6 answers => 14 Q's left and need 10 of these or more right to get an A

  5. Jack1
    • one year ago
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    therefore P of 10 out of 14 flips being correct

  6. Jack1
    • one year ago
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    = same as P of 10 out of 14 flips being heads

  7. Jack1
    • one year ago
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    =8.9% chance

  8. Raje
    • one year ago
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    perl why did you keep 9 instead of 10. your answer when given gives wrong result.

  9. gorv
    • one year ago
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    @perl it is wrong plzzz help me

  10. leena1996
    • one year ago
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    @gorv is it 28.6% because in 4 cases when there are 11,12,13&14 questions excluding the 6 initially correct are correct then 4/14 *100 % or 0.286 is the probability of her getting an A on the test.......this is a much easier way to get the answer...

  11. perl
    • one year ago
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    it is 2.86% , dont listen to leena

  12. perl
    • one year ago
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    let X = number of right answers among the 14 guessed questions. X is a binomial random variable , with n=14, p = .5 , X = 0,1,2,3... 14 She wants to get an A which is defined as obtaining more than 16 right answers (17 answers or greater). She already has 6 right answers, so she needs 11 or more guessed correctly. P( X >=11 ) = P( or X = 11 or X = 12 or X = 13 or X = 14) = P(X=11) + P(X=12) + P(X=13) + P( X = 14) where P(X=k) = nCk* p^k *(1-p)^(n-k) the TI 84 calculator can do it but using the complement P(X>=11) = 1 - P( X<11) = 1-P(X<=10) = 1 - binomcdf(14, .5, 10) = .0286865 = 2.86865%

  13. perl
    • one year ago
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    @gorv is it right?

  14. gorv
    • one year ago
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    yeppppppppppp thanku

  15. perl
    • one year ago
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    how many decimals do you round?

  16. leena1996
    • one year ago
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    @perl can u plz explain how is it 2.86?

  17. perl
    • one year ago
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    @leena1996 let X = number of right answers among the 14 guessed questions. X is a binomial random variable , with n=14, p = .5 , X = 0,1,2,3... 14 She wants to get an A which is defined as obtaining more than 16 right answers (17 answers or greater). She already has 6 right answers, so she needs 11 or more guessed correctly. P( X >=11 ) = P( or X = 11 or X = 12 or X = 13 or X = 14) = P(X=11) + P(X=12) + P(X=13) + P( X = 14) where P(X=k) = nCk* p^k *(1-p)^(n-k) the TI 84 calculator can do it but using the complement P(X>=11) = 1 - P( X<11) = 1-P(X<=10) = 1 - binomcdf(14, .5, 10) = .0286865 = 2.86865%

  18. perl
    • one year ago
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    i can explain more

  19. leena1996
    • one year ago
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    @perl i think there is some kind of catch here.....just check my reply and tell me whats wrong....

  20. perl
    • one year ago
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    ok so we have 14 questions we are looking at. right?

  21. gorv
    • one year ago
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    @leena1996 thankuuuuu very much

  22. perl
    • one year ago
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    she needs 11 or more right answers to get an A , (out of 14 questions)

  23. leena1996
    • one year ago
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    and plz do not put in Permutations and Combinations cuz we have only started with a little bit basic concepts in class

  24. perl
    • one year ago
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    uh oh :/

  25. perl
    • one year ago
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    ok i wont

  26. leena1996
    • one year ago
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    @gorv what is correct?? 2.86 or 28.6%

  27. leena1996
    • one year ago
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    @perl you may continue explaining your answer....

  28. perl
    • one year ago
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    ok so we have 14 questions we are looking at, and we want 11 or more right answers so its easier if we do the cases separately. when there is exactly 11 right , exactly 12 right, exactly 13 right , exactly 14 right

  29. perl
    • one year ago
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    case when exactly 11 right: here is one example where you have 11 right, 3 wrong SS SS SS SS SS SF FF

  30. perl
    • one year ago
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    we need 11 right or more to get an A , ok?

  31. perl
    • one year ago
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    S = success (right answer) F = fail (wrong answer )

  32. perl
    • one year ago
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    so imagine we order the questions from 1 - 14 , put them in a row. then to get 11 right you can have S S S S S S S S S S S F F F

  33. leena1996
    • one year ago
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    @perl this is exactly what i am saying see the conditions that satisfy success are 4 when 11,12,13 or 14 answers are correct. So 4 out of 14 are correct and so 4/14 = 0.286....why are u getting 0.0286?

  34. perl
    • one year ago
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    leena, yes but there are more cases S F S F S F S F ...

  35. gorv
    • one year ago
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    @leena1996 2.86 is correct

  36. perl
    • one year ago
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    for X = exactly 11 right SSSSSSSSSSFFF SFSFSFSSSSSS etc we can rearrange S's and F's,

  37. leena1996
    • one year ago
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    @Suchandrarao We are trying to figure out how it is the correct answer....ok.......and @perl i still dont get it....may be i will look into your explanation after we are taught Permutations and Combinations........alright thanks a lot!

  38. perl
    • one year ago
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    leena, wait

  39. perl
    • one year ago
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    lets look at a simpler problem you have a test with 4 questions, what is the probability of getting 3 right by simply guessing

  40. perl
    • one year ago
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    with me so far?

  41. leena1996
    • one year ago
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    ya i am with you @perl

  42. perl
    • one year ago
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    what are the possibilities. probability = # of favorable / total outcomes

  43. perl
    • one year ago
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    ok, so there are 4 questions , the possibities of getting exactly three right are SSSF SSFS SFSS FSSS

  44. perl
    • one year ago
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    SSSF = first three are right, last one is wrong SSFS = first two are right , third question is wrong, last question is right SFSS = first is right, 2nd is wrong, last two are right FSSS = first is wrong, last three are right

  45. leena1996
    • one year ago
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    @perl ya i get it but finally there are 3 favourable outcomes out of 4 so probability is 75% isnt it?

  46. leena1996
    • one year ago
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    @perl Plz explain

  47. perl
    • one year ago
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    no, because theres

  48. perl
    • one year ago
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    the *total* ways to get any right

  49. perl
    • one year ago
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    there are 4 favorable for exactly 3 right answers SSSF SSFS SFSS FSSS out of 16 total ways FFFF (none right) FFFS, FFSF, FSFF, SFFF (exactly 1 right) SSFF, SFSF, SFFS, FSFS, FSSF, FFSS (exactly 2 right) SSSF, SSFS, SFSS, FSSS (exactly 3 right) SSSS , (all right)

  50. perl
    • one year ago
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    4 favorable out of 16

  51. leena1996
    • one year ago
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    @perl ok so total ways are 16 out of which 4 are favourable......ya now i get it thanks a lot....i have learnt it i just had forgotton it and had no review in a lot of days........thanks for the explanation....

  52. perl
    • one year ago
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    but theres a way to do these without having to sit and write out all the cases

  53. perl
    • one year ago
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    ok, what about the probability of getting 3 or more right on this test?

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