Question

If two events A and B are such that P() = 0.3 P(2) = 0.4

and P(A Ē $\overline { \mathrm { B } }$ ) = 0.5 then P $\left( \frac { B } { A \cup \bar { B } } \right)$ =

Each of the following questions contains two statements; statement-1 and statement-2. Each of these question have four alternative choices, only one of which is correct answer. You have to select the correct choice

• A. 0.9
• B. 0.5
• C. 0.6
• D. 0.25

Answer 1

Answer: (D)

0.25

Answer 2

P= $\frac { \mathrm { P } [ \mathrm { B } \cap ( \mathrm { A } \cup \overline { \mathrm { B } } ) ] } { \mathrm { P } ( \mathrm { A } \cup \overline { \mathrm { B } } ) }$

=

$\left\{ \begin{array} { l } \mathrm { P } ( \mathrm { A } \cap \overline { \mathrm { B } } ) = 0.5 \\ \mathrm { P } ( \mathrm { A } ) - \mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) = 0.5 \\ \mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) = 0.7 - 0.5 = 0.2 \end{array} \right.$

= $\frac { \mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) } { \mathrm { P } ( \mathrm { A } ) + \mathrm { P } ( \overline { \mathrm { B } } ) - \mathrm { P } ( \mathrm { A } \cap \overline { \mathrm { B } } ) }$ = $\frac { 0.2 } { 0.7 + 0.6 - 0.5 }$

= $\frac { 0.2 } { 0.8 }$ = $\frac { 1 } { 4 }$ = 0.25