Posted by **noman** on Wednesday, March 6, 2013 at 6:42am.

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

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

- probability -
**saiko**, Wednesday, March 6, 2013 at 7:37am
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.

- probability -
**noman**, Wednesday, March 6, 2013 at 8:52am
thanx :) i got it

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