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 $\leq 5$.
State whether the following statement is true or false and provide a reason:
Statement: $A = B^{\prime}$
$A:$ 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 $\leq 5$.
State whether the following statement is true or false and provide a reason:
Statement: $A = B^{\prime}$
(A) The sample space $S$ for throwing two dice consists of $36$ outcomes.
Event $A$ is getting an even number on the first die:
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Event $B$ is getting an odd number on the first die:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
The complement of $B$,denoted as $B^{\prime}$,consists of all outcomes in $S$ that are not in $B$. Since the first die can only show an even or odd number,the complement of getting an odd number on the first die is getting an even number on the first die.
Therefore,$B^{\prime} = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Comparing the sets,we see that $A = B^{\prime}$.
Thus,the statement is true.
Event $A$ is getting an even number on the first die:
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Event $B$ is getting an odd number on the first die:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
The complement of $B$,denoted as $B^{\prime}$,consists of all outcomes in $S$ that are not in $B$. Since the first die can only show an even or odd number,the complement of getting an odd number on the first die is getting an even number on the first die.
Therefore,$B^{\prime} = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Comparing the sets,we see that $A = B^{\prime}$.
Thus,the statement is true.
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