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Home  >>  CBSE XII  >>  Math  >>  Probability
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Given that the events A and B are such that P(A) = \( \large \frac{1}{2} \) , P (A \( \cap\) B) = \( \large \frac{3}{5} \) and P(B) = p. Find p if they are (i) mutually exclusive (ii) independent.

$\begin{array}{1 1} (i) 1/10, (ii) 1/5 \\(i) 1/10, (ii) 2/5 \\ (i) 3/5, (ii) 2/5 \\ (i) 3/5, (ii) 1/5\end{array} $

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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.
Case (i): A and B are mutually exclusive.
Given \(\;P(A)=\large \frac{1}{2}, \;\) \(P(A\cup\;B)=\large \frac{3}{5},\)\(\;P(B)=p\).
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)
$\Rightarrow P (A \cap B) = 0 = \large\frac{1}{2} +$ $ p$$ - \large\frac{3}{5}$
$\Rightarrow p = \large\frac{3}{5}$$ - \large\frac{1}{2}$$ = \large \frac{1}{10}$
Case (ii): A and B are independent.
If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
P (A $\cap$ B) = P(A) + P(B) - P(A $\cup$ B)
$\Rightarrow$ P (A $\cap$ B) = P(A) + P(B) - P(A $\cup$ B) = $\large\frac{1}{2}$$ + p - $$ \large \frac{3}{5}$
$\Rightarrow$ P(A) $\times$ P(B) = $\large\frac{1}{2}$$ \times$ p
Solving, we get: $\large\frac{1}{2}$$ \times$ p = $\large\frac{1}{2}$$ + p - $$ \large \frac{3}{5}$
Solving for p, we get: $\large\frac{p}{2}$$ = \large \frac{3}{5} - \frac{1}{2} $ $=\large\frac{1}{10} \rightarrow$ $\;p = \large \frac{2}{10} $$= \large \frac{1}{5}$
answered Jun 19, 2013 by balaji.thirumalai
 

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