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Q. Two dice are thrown. The events $A$, $B$ and $C$ are as follows :
$A :$ getting an even number on the first die.
$B :$ getting an odd number on the first die.
$C :$ getting the sum of the numbers on the dice < 5.
Then, which of the following is true ?
(i) $A$ and $B$ are mutually exclusive.
(ii) $A$ and $B$ are mutually exclusive and exhaustive.
(iii) $A = B'$
(iv) $A$ and $C$ are mutually exclusive.
(v) $A$ and $B'$ are mutually exclusive.
(vi) $A'$, $B'$ and $C$ are mutually exclusive and exhaustive.

Probability

Solution:

$\because A =$ Getting an even number on the first die
$A = \{(2 , 1), (2, 2 ), (2, 3 ), (2. 4 ), (2, 5), (2 , 6 ), (4, 1)$,
$(4, 2), (4, 3 ), (4, 4), (4, 5), ( 4, 6), (6, 1), (6, 2), (6, 3), (6, 4)$,
$(6, 5), (6, 6)\}$
$B =$ Getting an odd number on the first die
$B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1)$,
$(3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2)$,
$(5 , 3), (5, 4) (5 , 5), (5 ,6)\}$
$C =$ Getting the sum of numbers $\le 5$
$= 1(1, 1), (1, 2), (1, 3 ), (1, 4), (2, 1), (2, 2), (2, 3)$,
$(3 ,1) , (3, 2), (4 ,1 )\}$
(i) True
$\Rightarrow A \cap B = \phi$
$\therefore A$ and $B$ are mutually exclusive events.
(ii) True
$\therefore A \cup B = S$ i.e., exhaustive. Also, $A \cap B = \phi$
(iii) True
$\therefore B =$ Getting an odd number on the first die
$\Rightarrow B' =$ Getting an even number on the first die $= A$
$\therefore A = B'$
(iv ) False
$\because A \cap C = \{(2 ,1 ), ( 2 ,2 ) , (2, 3 ), (4 ,1)\} \ne \phi$, so, $A$ and $C$ are not mutually exclusive.
(v ) False
$\because B' = A$
$\therefore A \cap B' = A \cap A = A \ne \phi$
So, $A$ and $B'$ are not mutually exclusive.
(vi) False
$\because A' \cap B' = \phi$
$\therefore A' \cap B' \cap C = \phi$ and $A' \cup B' \cup C = S$
But $A' \cap C = B \cap C$
$= \{(1, 1 ), (1, 2) , (1, 3), (1, 4), (3, 1), (3, 2)\} \ne \phi$ $(\because A' = B)$
and $B' \cap C = A \cap C = \{(2, 1), (2, 2), (2, 3), (4, 1)\} \ne \phi$
$(\because B' = A)$
$\therefore A'$, $B'$ and $C$ are not mutually exclusive.