# In a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ?

Here $n= 5$
p = probability of getting correct answer by giessing from 3 possible answers $=\frac{1}{3}$
$q= 1-p= \frac{2}{3}$
$\therefore p\left ( x\geq 4 \right )= p\left ( x= 4 \right )+p\left ( x= 5 \right )$
$= \, ^{5}C_{4}\, p^{4}\, q^{5-4}+\, ^{5}C_{5}\,\, p^{5}\, q^{5-5}$
$= 5\times \left ( \frac{1}{3} \right )^{4}\times \left ( \frac{2}{3} \right )^{1}+1\times \left ( \frac{1}{3} \right )^{5}\left ( \frac{2}{3} \right )^{0}$
$= \frac{10}{243}+\frac{1}{243}= \frac{11}{243}$

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