If E and F are events such that P(E)=frac{1}{4} , P(F)=frac{1}{2} and P(E and F )=frac{1}{8},

English

Gujarati

Hindi

14.Probability

easy

If $E$ and $F$ are events such that $P(E)=\frac{1}{4}$, $P(F)=\frac{1}{2}$ and $P(E$ and $F )=\frac{1}{8},$ find $:$ $P($ not $E$ and not $F)$.

$\frac{3}{8}$

Here, $P ( E )=\frac{1}{4}$, $P ( F )=\frac{1}{2},$ and $P ( E $ and $F )=\frac{1}{8}$

From $P ( E$ or $F )= P (E \cup F)=\frac{5}{8}$

We have $( E \cup F ) ^{\prime}=\left( E ^{\prime} \cap F ^{\prime}\right)$ $[$ By De Morgan's law $]$

$\therefore $ $( E ^{\prime} \cap F^{\prime})= P ( E \cup F ) ^{\prime}$

Now, $P ( E \cap F )^{\prime} =1- P ( E \cup F )$ $=1-\frac{5}{8}=\frac{3}{8}$

$\therefore $ $P(E^{\prime} \cap F^{\prime})=\frac{3}{8}$

Thus, $P($ not $E$ not $F)=\frac{3}{8}$

Standard 11

Mathematics

medium

easy

hard

(IIT-1988)

easy

easy

$P(A)$	$P(B)$	$P(A \cap B)$	$P (A \cup B)$
$\frac {1}{3}$	$\frac {1}{5}$	$\frac {1}{15}$	……..