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(C) The total number of possible outcomes when throwing $2$ dice is $6 	imes 6 = 36$.Let $A$ be the event of getting an odd number on the first die a

Vedclass · Accepted Answer

$\frac{11}{36}$

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(C) The total number of possible outcomes when throwing $2$ dice is $6 	imes 6 = 36$.Let $A$ be the event of getting an odd number on the first die and a multiple of $3$ on the second die.Odd numbers on a die are ${1, 3, 5}$. Multiples of $3$ on a die are ${3, 6}$.Possible outcomes for $(A, B)$ where $A \in {1, 3, 5}$ and $B \in {3, 6}$ are: $(1,3), (1,6), (3,3), (3,6), (5,3), (5,6)$. (Total $6$ outcomes).Let $C$ be the event of getting a multiple of $3$ on the first die and an odd number on the second die.Possible outcomes for $(C, D)$ where $C \in {3, 6}$ and $D \in {1, 3, 5}$ are: $(3,1), (3,3), (3,5), (6,1), (6,3), (6,5)$. (Total $6$ outcomes).The outcome $(3,3)$ is common to both sets.Using the inclusion-exclusion principle,the total number of unique favourable outcomes is $6 + 6 - 1 = 11$.The favourable outcomes are: $(1,3), (1,6), (3,3), (3,6), (5,3), (5,6), (3,1), (3,5), (6,1), (6,3), (6,5)$.Therefore,the required probability is $\frac{11}{36}$.

$2$ dice are thrown. Find the probability of getting an odd number on $1$ die and a multiple of $3$ on the other.

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