Question

The probabilities of a student getting I, II and III division in an examination are $\frac{1}{10},\frac{3}{5}\mathrm{and}\frac{1}{4}$ respectively. The probability that the student fails in the examination is
(a) $\frac{197}{200}$

(b) $\frac{27}{100}$

(c) $\frac{83}{100}$

(d) none of these

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\frac{27}{100} \\ P\left(\text{student gets first division}\right)=\frac{1}{10} \\ P\left(\text{student gets second division}\right)=\frac{3}{5} \\ P\left(\text{student gets third division}\right)=\frac{1}{4} \\ P\left(\text{students fail}\text{s}\right)=P\left(\text{student does not get first division}\right)\times P\left(\text{student does not get second division}\right)\times P\left(\text{student does not get third division}\right) \\ =\left(1-\frac{1}{10}\right)\left(1-\frac{3}{5}\right)\left(1-\frac{1}{4}\right) \\ =\frac{9}{10}\times \frac{2}{5}\times \frac{3}{4} \\ =\frac{54}{200} \\ =\frac{27}{100} \end{gathered}$$