A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

Solution :

Probability of guessing a correct answer, p = \(1\over 3\)

and probability of guessing a wrong answer, q  = \(2\over 3\)

So, the probability of guessing 4 or more correct answers is

= \(^5C_4\) \(({1\over 3})^4\). \(2\over 3\) + \(^5C_5\) \(({1\over 3})^5\)

= \(5.2\over {3^5}\) + \(1\over {3^5}\) = \(11\over {3^5}\)

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