The probabilities of a student getting I, II | vedclass

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(D) Let $A, B, C$ be the events of getting $I, II$ and $III$ division respectively,and $D$ be the event of failing the examination.Since these events

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $A, B, C$ be the events of getting $I, II$ and $III$ division respectively,and $D$ be the event of failing the examination.Since these events are mutually exclusive and exhaustive,the sum of their probabilities must be $1$.$P(A) = \frac{1}{10}, P(B) = \frac{3}{5}, P(C) = \frac{1}{4}$.$P(A) + P(B) + P(C) + P(D) = 1$$\frac{1}{10} + \frac{3}{5} + \frac{1}{4} + P(D) = 1$To add the fractions,find the least common multiple of $10, 5, 4$,which is $20$.$\frac{2}{20} + \frac{12}{20} + \frac{5}{20} + P(D) = 1$$\frac{19}{20} + P(D) = 1$$P(D) = 1 - \frac{19}{20} = \frac{1}{20}$.Since $\frac{1}{20} = 0.05$,which is not among the given options,the correct answer is $D$ (None of these).

Outcome	Probability
$\omega_{1}$	$\frac{1}{14}$
$\omega_{2}$	$\frac{2}{14}$
$\omega_{3}$	$\frac{3}{14}$
$\omega_{4}$	$\frac{4}{14}$
$\omega_{5}$	$\frac{5}{14}$
$\omega_{6}$	$\frac{6}{14}$
$\omega_{7}$	$\frac{15}{14}$

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

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