CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$A$$ and $$B$$ are two events such that $$P\left( A \right) = \displaystyle\frac { 1 }{ 4 } ,  P\left( B \right) = \displaystyle\frac { 1 }{ 2 } $$ and $$P\left( A\cap B \right) = \displaystyle\frac { 1 }{ 8 } $$, find $$8P\left( \bar A \  and\ \bar{B} \right) $$


Solution

It is given that, $$P\left( A \right) =\displaystyle\frac { 1 }{ 2 } $$ and $$P\left( A\cap B \right) = \displaystyle\frac { 1 }{ 8 } $$
$$P\left( not   on   A   and   not   on  B \right) =P\left( A\prime \cap B\prime  \right) $$
$$P\left( not   on   A   and   not   on   B \right) = P\left( \left( A\cup B \right)  \right) \prime   \left[ A\prime \cap B\prime = \left( A\cup B \right) \prime  \right] $$
$$=1 - P\left( A\cup B \right) $$
$$=1 - \left[ P\left( A \right) + P\left( B \right) - P\left( A\cap B \right)  \right] $$
$$= 1 - \left[ \displaystyle\frac { 1 }{ 4 } + \displaystyle\frac { 1 }{ 2 } - \displaystyle\frac { 1 }{ 8 }  \right] $$
$$= 1 - \displaystyle\frac { 5 }{ 8 } $$
$$= \displaystyle\frac { 3 }{ 8 }=0.27 $$

Mathematics
RS Agarwal
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image