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

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- schrodinger

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- gorv

if the coin lands Heads, she answers True, and if it lands Tails, she answers False

- Jack1

one point per question, yeah?

- Jack1

9%...?

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## More answers

- Jack1

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

- Jack1

therefore P of 10 out of 14 flips being correct

- Jack1

= same as P of 10 out of 14 flips being heads

- Jack1

=8.9% chance

- anonymous

perl why did you keep 9 instead of 10. your answer when given gives wrong result.

- gorv

@perl it is wrong plzzz help me

- anonymous

@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

it is 2.86% , dont listen to leena

- perl

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%

- perl

@gorv
is it right?

- gorv

yeppppppppppp thanku

- perl

how many decimals do you round?

- anonymous

@perl can u plz explain how is it 2.86?

- perl

@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%

- perl

i can explain more

- anonymous

@perl i think there is some kind of catch here.....just check my reply and tell me whats wrong....

- perl

ok so we have 14 questions we are looking at. right?

- gorv

@leena1996 thankuuuuu very much

- perl

she needs 11 or more right answers to get an A , (out of 14 questions)

- anonymous

and plz do not put in Permutations and Combinations cuz we have only started with a little bit basic concepts in class

- perl

uh oh :/

- perl

ok i wont

- anonymous

@gorv what is correct?? 2.86 or 28.6%

- anonymous

@perl you may continue explaining your answer....

- perl

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

- perl

case when exactly 11 right: here is one example where you have 11 right, 3 wrong
SS SS SS SS SS SF FF

- perl

we need 11 right or more to get an A , ok?

- perl

S = success (right answer)
F = fail (wrong answer )

- perl

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

- anonymous

@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

leena, yes but there are more cases
S F S F S F S F ...

- gorv

@leena1996 2.86 is correct

- perl

for X = exactly 11 right
SSSSSSSSSSFFF
SFSFSFSSSSSS
etc
we can rearrange S's and F's,

- anonymous

@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

leena, wait

- perl

lets look at a simpler problem
you have a test with 4 questions, what is the probability of getting 3 right by simply guessing

- perl

with me so far?

- anonymous

ya i am with you @perl

- perl

what are the possibilities.
probability = # of favorable / total outcomes

- perl

ok, so there are 4 questions , the possibities of getting exactly three right are
SSSF
SSFS
SFSS
FSSS

- perl

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

- anonymous

@perl ya i get it but finally there are 3 favourable outcomes out of 4 so probability is 75% isnt it?

- anonymous

@perl Plz explain

- perl

no, because theres

- perl

the *total* ways to get any right

- perl

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)

- perl

4 favorable out of 16

- anonymous

@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

but theres a way to do these without having to sit and write out all the cases

- perl

ok, what about the probability of getting 3 or more right on this test?

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