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On a multiple choice examination with three possibleanswers (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 correctanswers just by guessing? |
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Answer» 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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