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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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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if the coin lands Heads, she answers True, and if it lands Tails, she answers False
one point per question, yeah?
9%...?

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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
therefore P of 10 out of 14 flips being correct
= same as P of 10 out of 14 flips being heads
=8.9% chance
perl why did you keep 9 instead of 10. your answer when given gives wrong result.
@perl it is wrong plzzz help me
@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...
it is 2.86% , dont listen to leena
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%
@gorv is it right?
yeppppppppppp thanku
how many decimals do you round?
@perl can u plz explain how is it 2.86?
@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%
i can explain more
@perl i think there is some kind of catch here.....just check my reply and tell me whats wrong....
ok so we have 14 questions we are looking at. right?
@leena1996 thankuuuuu very much
she needs 11 or more right answers to get an A , (out of 14 questions)
and plz do not put in Permutations and Combinations cuz we have only started with a little bit basic concepts in class
uh oh :/
ok i wont
@gorv what is correct?? 2.86 or 28.6%
@perl you may continue explaining your answer....
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
case when exactly 11 right: here is one example where you have 11 right, 3 wrong SS SS SS SS SS SF FF
we need 11 right or more to get an A , ok?
S = success (right answer) F = fail (wrong answer )
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
@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?
leena, yes but there are more cases S F S F S F S F ...
@leena1996 2.86 is correct
for X = exactly 11 right SSSSSSSSSSFFF SFSFSFSSSSSS etc we can rearrange S's and F's,
@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!
leena, wait
lets look at a simpler problem you have a test with 4 questions, what is the probability of getting 3 right by simply guessing
with me so far?
ya i am with you @perl
what are the possibilities. probability = # of favorable / total outcomes
ok, so there are 4 questions , the possibities of getting exactly three right are SSSF SSFS SFSS FSSS
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
@perl ya i get it but finally there are 3 favourable outcomes out of 4 so probability is 75% isnt it?
@perl Plz explain
no, because theres
the *total* ways to get any right
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)
4 favorable out of 16
@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....
but theres a way to do these without having to sit and write out all the cases
ok, what about the probability of getting 3 or more right on this test?

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