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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?
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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gorv
 one year ago
Best ResponseYou've already chosen the best response.1if the coin lands Heads, she answers True, and if it lands Tails, she answers False

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1one point per question, yeah?

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1P 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

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1therefore P of 10 out of 14 flips being correct

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1= same as P of 10 out of 14 flips being heads

Raje
 one year ago
Best ResponseYou've already chosen the best response.1perl why did you keep 9 instead of 10. your answer when given gives wrong result.

gorv
 one year ago
Best ResponseYou've already chosen the best response.1@perl it is wrong plzzz help me

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@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...

perl
 one year ago
Best ResponseYou've already chosen the best response.3it is 2.86% , dont listen to leena

perl
 one year ago
Best ResponseYou've already chosen the best response.3let 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 *(1p)^(nk) the TI 84 calculator can do it but using the complement P(X>=11) = 1  P( X<11) = 1P(X<=10) = 1  binomcdf(14, .5, 10) = .0286865 = 2.86865%

perl
 one year ago
Best ResponseYou've already chosen the best response.3how many decimals do you round?

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@perl can u plz explain how is it 2.86?

perl
 one year ago
Best ResponseYou've already chosen the best response.3@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 *(1p)^(nk) the TI 84 calculator can do it but using the complement P(X>=11) = 1  P( X<11) = 1P(X<=10) = 1  binomcdf(14, .5, 10) = .0286865 = 2.86865%

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@perl i think there is some kind of catch here.....just check my reply and tell me whats wrong....

perl
 one year ago
Best ResponseYou've already chosen the best response.3ok so we have 14 questions we are looking at. right?

gorv
 one year ago
Best ResponseYou've already chosen the best response.1@leena1996 thankuuuuu very much

perl
 one year ago
Best ResponseYou've already chosen the best response.3she needs 11 or more right answers to get an A , (out of 14 questions)

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0and plz do not put in Permutations and Combinations cuz we have only started with a little bit basic concepts in class

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@gorv what is correct?? 2.86 or 28.6%

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@perl you may continue explaining your answer....

perl
 one year ago
Best ResponseYou've already chosen the best response.3ok 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

perl
 one year ago
Best ResponseYou've already chosen the best response.3case when exactly 11 right: here is one example where you have 11 right, 3 wrong SS SS SS SS SS SF FF

perl
 one year ago
Best ResponseYou've already chosen the best response.3we need 11 right or more to get an A , ok?

perl
 one year ago
Best ResponseYou've already chosen the best response.3S = success (right answer) F = fail (wrong answer )

perl
 one year ago
Best ResponseYou've already chosen the best response.3so 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

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@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?

perl
 one year ago
Best ResponseYou've already chosen the best response.3leena, yes but there are more cases S F S F S F S F ...

gorv
 one year ago
Best ResponseYou've already chosen the best response.1@leena1996 2.86 is correct

perl
 one year ago
Best ResponseYou've already chosen the best response.3for X = exactly 11 right SSSSSSSSSSFFF SFSFSFSSSSSS etc we can rearrange S's and F's,

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@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!

perl
 one year ago
Best ResponseYou've already chosen the best response.3lets look at a simpler problem you have a test with 4 questions, what is the probability of getting 3 right by simply guessing

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0ya i am with you @perl

perl
 one year ago
Best ResponseYou've already chosen the best response.3what are the possibilities. probability = # of favorable / total outcomes

perl
 one year ago
Best ResponseYou've already chosen the best response.3ok, so there are 4 questions , the possibities of getting exactly three right are SSSF SSFS SFSS FSSS

perl
 one year ago
Best ResponseYou've already chosen the best response.3SSSF = 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

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@perl ya i get it but finally there are 3 favourable outcomes out of 4 so probability is 75% isnt it?

perl
 one year ago
Best ResponseYou've already chosen the best response.3the *total* ways to get any right

perl
 one year ago
Best ResponseYou've already chosen the best response.3there 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)

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@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....

perl
 one year ago
Best ResponseYou've already chosen the best response.3but theres a way to do these without having to sit and write out all the cases

perl
 one year ago
Best ResponseYou've already chosen the best response.3ok, what about the probability of getting 3 or more right on this test?
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