On a multiple choice examinations with 4 possible answer (out of which one is correct) for each of the five question. The probability that a candidate would get 4 or more correct answer just by guessing is :

$\frac{1}{64}$

The correct answer is Option (1) → $\frac{1}{64}$

P(success) = $\frac{1}{4}$, P(failure) = $\frac{3}{4}$

P(4 or more correct) = P(4 success) + P(5 success)

$={^5C}_4(\frac{1}{4})^4(\frac{3}{4})+{^5C}_5(\frac{1}{4})^5$

$=\frac{5×3}{4^5}+\frac{1}{4^5}=\frac{16}{4^5}=\frac{1}{4^3}=\frac{1}{64}$