On a multiple-choice exam with 3 possible answers for each of the 5 questions, find the probability that a student would get 4 or more correct answers just by guessing.
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OpenStudy (idealist10):
@ganeshie8 @wio @radar
ganeshie8 (ganeshie8):
binomial distribution ?
OpenStudy (idealist10):
I think so.
ganeshie8 (ganeshie8):
just addup the probabilities p(X>=4) = P(4) + P(5)
ganeshie8 (ganeshie8):
start by finding the probabiltiy for sucess in guessing a correct answer
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OpenStudy (idealist10):
So is it P(4)+P(5)=(1/3)^4+(1/3)^5?
ganeshie8 (ganeshie8):
thats a good try, but no
P(5) equals (1/3)^5
P(4) is bit tricky as you have a choice to tick false option for any one of the five questions
ganeshie8 (ganeshie8):
S : success
F : failure
ganeshie8 (ganeshie8):
you can answer exactly four correct answers in any one of below 5 ways
```
FSSSS
SFSSS
SSFSS
SSSFS
SSSSF
```
yes ?
OpenStudy (idealist10):
Yes. And then?
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ganeshie8 (ganeshie8):
Notice each of them have four success ticks and one failure tick
so the probability of each of them is same : (1/3)^4 * (2/3)^1
ganeshie8 (ganeshie8):
P(4) = 5 * (1/3)^4 * (2/3)^1
ganeshie8 (ganeshie8):
see if that makes sense more or less ^
OpenStudy (idealist10):
Yes, it makes sense! Thanks!
ganeshie8 (ganeshie8):
yw :) add that to P(5) :
P(X>=4) = P(4) + P(5)
= 5 * (1/3)^4 * (2/3)^1 + (1/3)^5
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