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Home  >>  CBSE XII  >>  Math  >>  Probability

If A and B are two events such that P (A) = \(\large \frac{1}{4} \) , P (B) = \( \large\frac{1}{2} \) and P(A \(\cap\)B) = \(\large \frac{1}{8} \) , find P (not A and not B).

$\begin{array}{1 1}3/8 \\ 1/8 \\ 1/4 \\ 5/8 \end{array} $

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  • P (not A and not B) = \(P(\bar{A}\cap\;\bar{B})\) = \(P(\overline{A \cup B})\) = 1 - P (A $\cup$ B)
  • P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B)
  • P (\(\bar{A})\) = 1 - P(A)
Given \(\;P(A)=\large \frac{1}{4}\), \(P(B)=\large \frac{1}{2}\) and \(\;P(A\cap\;B)=\large \frac{1}{8}\)
P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B)
\(\Rightarrow \;P(A\cup\;B)=\large \frac{1}{4}+\frac{1}{2}-\frac{1}{8}\) =\(\large\;\frac{5}{8}\)
\(\Rightarrow P(\bar{A}\cap\;\bar{B})=1-P(A\cup\;B)\) =\(\;1-\large\frac{5}{8}=\frac{3}{8}\)
answered Jun 19, 2013 by balaji.thirumalai
 

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