What is the probability of getting 80% or more of the questions correct on a 10-question true-false exam merely by guessing?

Question

probability - Damon, Wednesday, March 6, 2013 at 2:53am

Binomial distribution with p = .5

p(10 right)
= c(10,10) .5^10 (.5)^0 =.5^10 = .000976

p( 9 right)
= c(10,9) .5^9 (.5)^1 = .00976

p(8 right)
= c(10,8) (.5)^8 (.5)^2 = 45*.5^10 = .04394

sum = .0547


I can't understand that way.... can anyone explain or solve it another way

Answer 1

well you can understand it like,
for one qusetion getting right,
the probability is=.5
as there are only two options true and false.
for two qusetion getting right,
the probability is=.5*.5
for three qusetion getting right,
the probability is=.5*.5*.5
.................................................................
for eight qusetion getting right,
the probability is=(.5)^8
for exactly eight question getting right you have to make 2 questions wrong.
for two qusetion getting wrong,
the probability is=(.5)^2
and you can select 8 questions out of 10 at a probability=C(10,10)*(.5)^8*(.5)^2
.
.
.
thus you can calculate probability for exactly 9 qstn and 10 qstn right.
sum them and u'ld get the ans.

Answer 2

thanx :) i got it