Question

If two events A and B are such that

P(1) = 0.3 ; P(2) = 0.4 ; P$( \overline { \mathrm { A } } \cap \overline { \mathrm { B } } )$= 0.5 then $P \left( \frac { B } { A \cup \bar { B } } \right)$ =

• A. ¾
• B. 5/6
• C. ¼
• D. 3/7

Answer 1

Answer: (C)

¼

Answer 2

P $\left( \frac { B } { A \cup \bar { B } } \right)$ = $\frac { \mathrm { P } ( \mathrm { B } \cap ( \mathrm { A } \cup \overline { \mathrm { B } } ) ) } { \mathrm { P } ( \mathrm { A } \cup \overline { \mathrm { B } } ) }$

{ B Ç (A È ) = (B Ç A) È (B Ç ) &B Ç = f}

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

= $\frac { \mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) } { \mathrm { P } ( \mathrm { A } ) + \mathrm { P } ( \overline { \mathrm { B } } ) - \mathrm { P } ( \mathrm { A } \cap \overline { \mathrm { B } } ) }$ = $\frac { 0.2 } { 0.3 + 0.6 - 0.1 }$

= $\frac { 0.2 } { 0.8 }$ = ¼

P $( \overline { \mathrm { A } } \cap \overline { \mathrm { B } } )$ = 0.5 ̃ 1 – P(A È B) = 0.5

P (A È B) = 0.5 ̃ P(1)+P(2)–P(A Ç B) = 0.5

0.3 + 0.4 – x = 0.5 ̃ P(A Ç B) = x = 0.7 – 0.5

P(AÇB)=x=0.2̃P(A Ç ) = P(1) – P(AÇ B)

= 0.3 – 0.2 = 0.1