Question

14. If E and F are events such that P ( E ) = 1/4 , P ( F ) = 1/2 and P ( E and F)=1/8 find P ( not E and not F ).

Answer 1

$\large\underline{\sf{Given- }}$

$\boxed{ \sf \: P(E) = \dfrac{1}{4} } \: and \: \boxed{ \sf \: P(F) = \dfrac{1}{2} } \: and \: \boxed{ \sf \: P(E \cap \: F) = \dfrac{1}{8} }$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{P(not \: E \: and \: not \: F)}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \sf \: P(E \: \cup \: F) = P(E) + P(F) - P(E \cap \: F) }$

$\boxed{ \sf \: P( \overline{E} \: \cap \: \overline{F}) =1 - P(E \: \cup \: F) }$

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{P(E) = \dfrac{1}{4} } \\ &\sf{P(F) = \dfrac{1}{2} }\\ &\sf{P(E \: \cap \: F) = \dfrac{1}{8} } \end{cases}\end{gathered}\end{gathered}$

We know that,

$\rm :\longmapsto\:P(E \: \cup \: F) = P(E) + P(F) - P(E \cap \: F)$

On substituting all the given values, we get

$\rm :\longmapsto\:P(E \: \cup \: F) = \dfrac{1}{4} + \dfrac{1}{2} - \dfrac{1}{8}$

$\rm :\longmapsto\:P(E \: \cup \: F) = \dfrac{2 + 4 - 1}{8}$

$\rm :\longmapsto\:P(E \: \cup \: F) = \dfrac{5}{8}$

Now,

we have to find the value of P( not E and not F )

That is

$\rm :\longmapsto\:P(notE \: and \: notF)$

$\sf \: = \: P(\overline{E} \: \cap \: \: \overline{F})$

$\sf \: = \: 1 \: - \: \:P(E \: \cup \: F)$

$\sf \: = \: 1 - \dfrac{5}{8}$

$\sf \: = \: \dfrac{3}{8}$

$\overbrace{ \underline { \boxed { \rm \therefore \: P(not \: E \: and \: not \: F) = \: \dfrac{3}{8} }}}$

Additional Information :-

$\boxed{ \sf \: P(\overline{E}) =1 - P(E) }$

$\boxed{ \sf \: P(E \: \cap \: \overline{F}) = P(E) - P(E \: \cap \: F)}$

$\boxed{ \sf \: P(F \: \cap \: \overline{E}) = P(F) - P(E \: \cap \: F)}$

$\boxed{ \sf P( \overline{E} \: \cup \: \overline{F}) =1 - P(E \: \cap \: F)}$