Given, In a multiple choice test there are five alternate options for first two question and 4 alternate options for third question.
The probability to do first question correctly, $\mathrm{P}\left(\mathrm{F}1\mathrm{C}\right)=\frac{1}{5}$
The probability to do second question correctly, $\mathrm{P}\left(\mathrm{F}2\mathrm{C}\right)=\frac{1}{5}$
The probability to do third question correctly, $\mathrm{P}\left(\mathrm{F}3\mathrm{C}\right)=\frac{1}{4}$
Similarly,The probability to do first question wrong, $\mathrm{P}\left(\mathrm{F}1\mathrm{U}\right)=\frac{4}{5}$
The probability to do second question wrong, $\mathrm{P}\left(\mathrm{F}2\mathrm{U}\right)=\frac{4}{5}$
The probability to do third question wrong, $\mathrm{P}\left(\mathrm{F}3\mathrm{U}\right)=\frac{3}{4}$
The probability to do exactly one question correctly=(Probability to do first question correct and rest wrong)+(Probability to do second question correct and rest wrong)+(Probability to do third question correct and rest wrong)

= (P(F1C) × P(F2U) × P(F3U)) + (P(F2C) × P(F1U) × P(F3U)) + (P(F3C) × P(F2U) × P(F1U))

$=\left(\frac{1}{5}\times \frac{4}{5}\times \frac{3}{4}\right)+\left(\frac{1}{5}\times \frac{4}{5}\times \frac{3}{4}\right)+\left(\frac{1}{4}\times \frac{4}{5}\times \frac{4}{5}\right)$

$\mathrm{P}=\frac{(3+3+4)}{25}=\frac{2}{5}$