Question

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 $\le 5$.
Then, which of following is not correct?
(i) $A^{\prime }=B$
(ii) Not $B=A^{\prime }$
(iii) $A$ or $B=A\cup B=$ All elements of $A$ and $B$
(iv) $A$ and $B=\varphi$
(v) $A$ but not $C=\{(2,4),(2,5),(2,6),(4,2),(4,3)$, $(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5)$, $(6,6)\}$
(vi) $B$ or $C=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$, $(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5$, $2),(5,3),(5,4),(5,5),(5,6),(2,1),(2,2),(2,3)$, $(4,1)\}$
(vii) $B$ and $C=\{(1,1),(1,2),(1,3),(14),(3,1),(3,2)\}$
(viii)$A\cap B^{\prime }\cap C^{\prime }=\{(2,4),(2,5),(2,6),(4,2),(4,3)(4,4)(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5)(6,6)\}$

• A. (ii)
• B. (i), (ii), ( (iii)
• C. (iii), (iv), (v)
• D. All of these

Answer 1

Answer: (A)

If two dice are thrown then, total number of possible outcomes, $S=6\times 6=36$ which are as follows
$A=$ Getting an even number on the first die
$=\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1)$,
$(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2)$,
$(6,3),(6,4),(6,5),(6,6\}$
$B=$ Getting an odd number on the first die
$=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1)$,
$(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2)$.
$(5,3),(5,4),(5,5),(5,6)\}$
$C=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3)$,
$(3,1),(3,2),(4,1)\}$
(i) $A^{\prime }=$ Which are not in $A=B$ (if $A$ is an event getting an even number on the first die, so $A^{\prime }$ will be the event getting an odd number on the first die i.e., $B$ )
(ii) not $B=B^{\prime }=$ which are not in $B=A$ ( $B$ is an event getting an odd number on the first, so $B^{\prime }$ will be the event getting an even number on the first die i.e., $A$ )
(iii) $A$ or $B=A\cup B$
$A\cup B$ includes all elements of $A$ and $B$ i.e., it will includes both the events $A$ and $B$ i.e., event having an even number on the first die and event having odd number on the first die i.e., $A\cup B$ will include all possible event i.e., $S$ (sample space)
$\therefore A\cup B=S$
(iv) $A$ and $B=A\cap B$ i.e., elements which are common in both $A$ and $B$. There is no element common in $A$ and $B$.
So, $A\cap B=\varphi$ (null set)
(v) $A$ but not $C=A-C=$ Elements which are in $A$ but not in C.
$=\{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),$
$(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
(vi) $B$ or $C=B\cup$ Ci.e., elements which are in both $B$ or $C$.
$=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),$
$(5,3),(5,4),(5,5),(5,6),(2,1),(2,2),(2,3),(4,1)\}$
(vii) $B$ and $C=B\cap C$ i.e., elements which are common in both $B$ and $C$.
$=\{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\}$
(viii) $A\cap B^{\prime }\cap C^{\prime }=\{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4$, 5). $(4,6).(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$