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Question
The probabilities of a student getting I, II and III division in an examination are \[\frac{1}{10}, \frac{3}{5}\text{ and } \frac{1}{4}\]respectively. The probability that the student fails in the examination is
Options
\[\frac{197}{200}\]
\[\frac{27}{100}\]
\[\frac{83}{100}\]
none of these
Solution
\[ \frac{27}{100}\]
\[P\left( \text{ student gets first division } \right) = \frac{1}{10}\]
\[P\left( \text{student gets second division } \right) = \frac{3}{5}\]
\[P\left( \text{ student gets third division } \right) = \frac{1}{4}\]
\[P\left( \text{ students fails } \right) = P\left( \text{ student does not get first division } \right) \times P\left( \text{ student does not get second division } \right) \times P\left( \text{ student does not get third division } \right)\]
\[ = \left( 1 - \frac{1}{10} \right)\left( 1 - \frac{3}{5} \right)\left( 1 - \frac{1}{4} \right)\]
\[ = \frac{9}{10} \times \frac{2}{5} \times \frac{3}{4}\]
\[ = \frac{54}{200}\]
\[ = \frac{27}{100}\]
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