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 :

Engineering

Mathematics

Binomial Probability Theorem

Question

$\frac{10}{3^{5}}$

$\frac{13}{3^{5}}$

$\frac{17}{3^{5}}$

$\frac{11}{3^{5}}$

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Favourable case = 4 correct + 5 correct

$=^{5} C_{4} {(\frac{1}{3})}^{4} (\frac{2}{3}) + {(\frac{1}{3})}^{5}$

$= 10 {(\frac{1}{3})}^{5} + {(\frac{1}{3})}^{5}$

$= \frac{11}{3^{5}}$

Aliter: n(S) = 3⁵; n(A) = ⁵C₄ × 2 + ⁵C₅

Required probability $= \frac{^{5} C_{4} \times 2 + 1}{3^{5}}$