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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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Let probability of getting a correct answer is `P(A) `.
Here, `P(A) = 1/3`
`:. P(barA) = 1-1/3 = 2/3`
Here, `P(barA)` is the probability of getting an incorrect answer.
Now, there are two cases when a candidate can answer four or more answers correctly.
Case 1: When he answers 4 questions correcly and one incorrect.
In this case, probability will be `= C(5,4)(1/3)^4(2/3) = 5*1/81*1/3 = 10/243`
Case 2: When he answers all 5 questions correcly .
In this case, probability will be `= C(5,5)(1/3)^5 = 1*1/243 = 1/243`
`:.` Required probaility ` = 10/243+1/243 = 11/243.`

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