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13. probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths13. probability1 Mark
MCQ
Consider two events $A$ and $B$ such that $P(A) = \frac{1}{4},\,\,P\left( {\frac{B}{A}} \right) = \frac{1}{2},\,\,P\left( {\frac{A}{B}} \right) = \frac{1}{4}.$ For each of the following statements, which is true

$I.$    $P\,({A^c}/{B^c}) = \frac{3}{4}$

$II.$   The events $A$ and $B$ are mutually exclusive

$III.$  $P(A/B) + P(A/{B^c}) = 1$

  • ✓
    $I$ only
  • B
    $I$ and $II$
  • C
    $I$ and $III$
  • D
    $II$ and $III$
✓

Answer

Correct option: A.
$I$ only
a
(a) $P\left( {\frac{B}{A}} \right) = \frac{{P(A \cap B)}}{{P(A)}}\, \Rightarrow \,\frac{1}{2} = \frac{{P(A \cap B)}}{{1/4}}$

$ \Rightarrow P(A \cap B) = \frac{1}{8}$

Hence events $A$ and $B$ are not mutually exclusive.

$\therefore$ Statement $II$ is incorrect.

$P\,\left( {\frac{A}{B}} \right) = \frac{{P(A \cap B)}}{{P(B)\,}} \Rightarrow \,P(B) = \frac{1}{2}$

$\because$ $P(A \cap B) = \frac{1}{8} = P(A)\,.\,P(B)$

$\therefore$ events $A$ and $B$ are independent events.

$P\,\left( {\frac{{{A^c}}}{{{B^c}}}} \right) = \frac{{P({A^c} \cap {B^c})}}{{P({B^c})}} = \frac{{P({A^c})\,P({B^c})}}{{P\,({B^c})}} $

$= \frac{3}{4}.\,\frac{1}{2}.\,\frac{2}{1} = \frac{3}{4}$

Hence statement $I$ is correct.

Again $P\left( {\frac{A}{B}} \right) + \,P\left( {\frac{A}{{{B^c}}}} \right)$

$ = \frac{1}{4} + \frac{{P(A \cap {B^c})}}{{P({B^c})}}$

$ = \frac{1}{4} + \frac{{P(A) - P(A \cap B)}}{{P({B^c})}}$

$ = \frac{1}{4} + \frac{{\frac{1}{4} - \frac{1}{8}}}{{\frac{1}{2}}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Hence statement $III$ is incorrect.

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