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Home  >>  CBSE XI  >>  Math  >>  Probability
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Two dice are thrown.The events A,B and C are as follows :A : getting an even number in first die,B : getting an odd number on the first die,C : getting the sum of the numbers on the dice $\leq$ 5.State true or false : A and C are mutually exclusive

$\begin{array}{1 1}(A)\;\text{True}\\(B)\;\text{False}\end{array} $

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1 Answer

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Given two dice are thrown
Possible outcomes =$6\times 6=36$
$\therefore$ The sample space
$S=\left\{\begin{array}{1 1}(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)\\(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)\\(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)\\(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)\\(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)\\(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\end{array}\right\}$
A: Getting an even number on first die
A=$\left\{\begin{array}{1 1}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}$
C : Getting the sum of the numbers on the dice $\leq 5$
$C=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\}$
$A\cap C\neq \phi$
There are few common elements b/w A and C
$\therefore A \cap C=\{(2,1),(2,2),(2,3),(4,1)\}\neq \phi$
$\therefore A $ and C are not mutually exclusive.
Hence the given statement is False.
answered Jun 30, 2014 by sreemathi.v
 

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