## gorv 2 years 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?

1. gorv

2. Jack1

one point per question, yeah?

3. Jack1

9%...?

4. 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

5. Jack1

therefore P of 10 out of 14 flips being correct

6. Jack1

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

7. Jack1

=8.9% chance

8. Raje

9. gorv

@perl it is wrong plzzz help me

10. leena1996

@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

it is 2.86% , dont listen to leena

12. 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%

13. perl

@gorv is it right?

14. gorv

yeppppppppppp thanku

15. perl

how many decimals do you round?

16. leena1996

@perl can u plz explain how is it 2.86?

17. 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%

18. perl

i can explain more

19. leena1996

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

20. perl

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

21. gorv

@leena1996 thankuuuuu very much

22. perl

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

23. leena1996

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

24. perl

uh oh :/

25. perl

ok i wont

26. leena1996

@gorv what is correct?? 2.86 or 28.6%

27. leena1996

28. 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

29. perl

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

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

31. perl

32. 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

33. leena1996

@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

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

35. gorv

@leena1996 2.86 is correct

36. perl

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

37. leena1996

@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

leena, wait

39. 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

40. perl

with me so far?

41. leena1996

ya i am with you @perl

42. perl

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

43. perl

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

44. 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

45. leena1996

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

46. leena1996

@perl Plz explain

47. perl

no, because theres

48. perl

the *total* ways to get any right

49. 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)

50. perl

4 favorable out of 16

51. leena1996

@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

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

53. perl

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