Probabilities of a student getting I, II and III division in an examination are respectively frac{1}

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14.Probability

medium

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

A

$\frac{{197}}{{200}}$

B

$\frac{{27}}{{100}}$

C

$\frac{{83}}{{100}}$

D

None of these

Solution

(d) $A$ denote the event getting $I$;

$B$ denote the event getting $II$;

$C$ denote the event getting $III$;

and $D$ denote the event getting fail.

Obviously, these four event are mutually exclusive and exhaustive,

therefore $P(A) + P(B) + P(C) + P(D) = 1 $

$\Rightarrow P(D) = 1 – 0.95 = 0.05$.

Standard 11

Mathematics

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