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 : $P ( E$ or $F )$

A

$\frac{5}{8}$

B

$\frac{5}{8}$

C

$\frac{5}{8}$

D

$\frac{5}{8}$

Solution

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

We know that $P ( E$ and $F )= P ( E )+ P ( F )- P ( E$ and $F )$

$\therefore P(E $ or $F)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$ $=\frac{2+4-1}{8}=\frac{5}{8}$

Standard 11

Mathematics

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