✓
Answer
Two balanced dice are thrown simultaneously.
$\therefore n = 62 = 36$
A = Event that the sum of numbers on two dice is a multiple of $2$ i.e., $2, 4, 6, 8, 10$ or $12.$
$= \{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\}$
$\therefore$ Favourable outcomes for the event $A$ is
$m = 18.\therefore P(A)=\frac{m}{n}=\frac{18}{36}$
$B =$ Event that the sum of numbers on the dice is a multiple of $3$ i.e., $3, 6, 9$ or $12.$
$= \{(1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)\}$
$\therefore$ Favourable outcomes for the event $B$ is
$m = 12.$
$\therefore p(B)=\frac{m}{n}=\frac{12}{36}$
$A \cap B =$ Event that the sum of the numbers on dice is a multiple of $2$ and $3,$ i.e.,$ 6$ or $12.$
$= \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)\}$
$\therefore$ Favourable outcomes for the event $A \cap B$ is $m = 6.$
$\therefore P(A \cap B)=\frac{m}{n}=\frac{6}{36}$
Now, $A \cup B =$ Event that the sum of numbers on dice is a multiple of $2$ or $3.$
$\therefore P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$=\frac{18}{36}+\frac{12}{36}-\frac{6}{36}$
$=\frac{24}{36}=\frac{2}{3}$
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