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},

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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 $F)$.

A

$\frac{3}{8}$

B

$\frac{3}{8}$

C

$\frac{3}{8}$

D

$\frac{3}{8}$

Solution

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

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