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If E and F are events .Such that $P(E)=\large\frac{1}{4}$$,P(F)=\large\frac{1}{2}$ and P(E and F)=$\large\frac{1}{8}$.Find P(not E and not F)

$\begin{array}{1 1}(A)\;\large\frac{3}{8}\\(B)\;\large\frac{7}{8}\\(C)\;\large\frac{5}{8}\\(D)\;\large\frac{1}{8}\end{array} $

1 Answer

  • $P(E \cup F)=P(E)+P(F)-P(E \cap F)$
P(not E and not F)=P(E' and F')=$P(E'\cap F')$
According to Demorgan's law
$P(E' \cap F')=P(E \cup F)'$
$\therefore P(E \cup F)'=1-P(E \cup F)$
$\Rightarrow 1-\large\frac{5}{8}$
$\Rightarrow \large\frac{3}{8}$
$\therefore$ P(not E and not F)=$\large\frac{3}{8}$
Hence (A) is the correct answer.
answered Jul 1, 2014 by sreemathi.v

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