Question
Imagine taking a 10-question true or false exam. You randomly guess at each question. Don't do any calculations, just tell me your first impressions
a). Is this a binomial setting?
b) How many question do you think you would get correct?
c). How surprised would you be if you passed ( 6 out of 10 correct or better) the exam?
d) How surprise would you be if you got an "A" on the exam (9 or 10 out of 10)?
a). Is this a binomial setting?
b) How many question do you think you would get correct?
c). How surprised would you be if you passed ( 6 out of 10 correct or better) the exam?
d) How surprise would you be if you got an "A" on the exam (9 or 10 out of 10)?
Answers
Damon
ell, I will just do the problem
a yes
p(right) = .5, (1 -pright)=.5
b half of them, 5
10 trials
p(k) = C(n,k) p^k (1-p)^(n-k)
p(6)=C(10,6) .5^6 .5^4 =210*.015625*.0625 = .205
p(7) = 120*.5^7*.5^3 = .117
p(8) = 45*.5^8 *.5^2 = .044
p(9) = 10*.5^9*.5 = .0098
p(10) = 10*.5^10 = .0005
so prob of six through ten = sum = .376 is probability of passing
p(9)+p(10) = .0005+.044 = .045
a yes
p(right) = .5, (1 -pright)=.5
b half of them, 5
10 trials
p(k) = C(n,k) p^k (1-p)^(n-k)
p(6)=C(10,6) .5^6 .5^4 =210*.015625*.0625 = .205
p(7) = 120*.5^7*.5^3 = .117
p(8) = 45*.5^8 *.5^2 = .044
p(9) = 10*.5^9*.5 = .0098
p(10) = 10*.5^10 = .0005
so prob of six through ten = sum = .376 is probability of passing
p(9)+p(10) = .0005+.044 = .045