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Home  >>  CBSE XII  >>  Math  >>  Probability
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Let A and B be independent events with P (A) = 0.3 and P(B) = 0.4. Find (i) P(A $\cap$ B) (ii) P(A $\cup$ B) (iii) P ($\large\frac{A}{B}$) (iv) P ($\large\frac{B}{A}$)

$\begin{array}{1 1} (i) 0.12 \quad (ii) 0.58 \quad (iii) 0.3 \quad (iv) 0.4 \\(i) 0.24 \quad (ii) 0.68 \quad (iii) 0.4 \quad (iv) 0.3\\ (i) 0.24 \quad (ii) 0.58 \quad (iii) 0.4 \quad (iv) 0.3 \\ (i) 0.16 \quad (ii) 0.68 \quad (iii) 0.2 \quad (iv) 0.8 \end{array} $

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1 Answer

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Toolbox:
  • If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
  • If A and B are mutually exclusive, then P (A $\cap$ B) = 0.
  • P (A $\cap$ B) = P(A) + P(B) - P(A $\cup$ B)
  • \(\;P(B/A)=\large \frac{P(A\cap\:B)}{P(A)}\)\(\:;P(A/B)=\large \frac{P(A\cap\;B)}{P(B)}\)
(i) Calculate P (A$\cap$ B):
If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
$\Rightarrow$ \(P(A\cap\;B)=P(A)\;P(B)\) = 0.3 $\times$ 0.4 = 0.12.
(ii) Calculate P (A $\cup$ B):
P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B)
$\Rightarrow$ P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B) = 0.3+0.4 - 0.12 = 0.58
(iii) Calculate P(A/B):
\(P(A/B)=\large \frac{P(A\cap\;B)}{P(B)}\)
$\Rightarrow$ P(A/B)=\(\large \frac{P(A\cap\;B)}{P(B)}\) = $\large\frac{0.12}{0.4}$ = 0.3
(iv) Calculate P(B/A):
\(P(B/A)=\large \frac{P(A\cap\;B)}{P(A)}\)
$\Rightarrow$ P(B/A)=\(\large \frac{P(A\cap\;B)}{P(A)}\) = $\large\frac{0.12}{0.3}$ = 0.4
answered Jun 19, 2013 by balaji.thirumalai
 

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