Question

The probabilities of a student getting I, II, and III division in an examination are $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.,$, and $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ respectively.

The probability that the student fails in the examination is

• A. $\raisebox{1ex}{$197$}\!\left/ \!\raisebox{-1ex}{$200$}\right.$
• B. $\raisebox{1ex}{$27$}\!\left/ \!\raisebox{-1ex}{$100$}\right.$
• C. $\raisebox{1ex}{$83$}\!\left/ \!\raisebox{-1ex}{$100$}\right.$
• D. None of these

Answer 1

The correct option is B

$\raisebox{1ex}{$27$}\!\left/ \!\raisebox{-1ex}{$100$}\right.$

Explanation for the correct option

The probability of a student getting I division, $P\left(A\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$

The probability of a student getting II division, $P\left(B\right)=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

The probability of a student getting III division, $P\left(C\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

The probability $P$ that the student fails the exam can be given by: $\left(\text{Probability of not getting I division}\right)\left(\text{Probability of not getting II division}\right)\left(\text{Probability of not getting III division}\right)$.

$$\begin{gathered} P=\left[1-\left(\frac{1}{10}\right)\right]\left[1-\frac{3}{5}\right]\left[1-\frac{1}{4}\right] \\ \Rightarrow P=\left(\frac{9}{10}\right)\left(\frac{2}{5}\right)\left(\frac{3}{4}\right) \\ \Rightarrow P=\frac{27}{100} \end{gathered}$$

Therefore, the required probability is $\frac{27}{100}$.

Hence option (B) is the correct option.