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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 - ,

    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 - ,

    thanx :) i got it

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