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On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

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The correct Answer is:
N/a
Here `n=5`
`p=` probability of correct answer of 1 question
`=1/3`
`q=1-p=1-1/3=2/3`
Now, the probability of correct answer of 4 or more than 4 questions.
`=P(Xge4)=P(X=4)+P(X=5)`
`=.^(5)C_(4).p^(4).q^(1)+.^(5)C_(5).p^(5)`
`=5xx(1/3)^(4) . 2/3+1xx(1/3)^(5)`
`=(10+1)/243=11/243`
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