Twelve tickets are numbered 1 to 12. One tick | vedclass

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Question

(A) The total number of outcomes is $n(S) = 12$. Let $A$ be the event that the number is divisible by $2$. The numbers are ${2, 4, 6, 8, 10, 12}$,so $

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) The total number of outcomes is $n(S) = 12$. Let $A$ be the event that the number is divisible by $2$. The numbers are ${2, 4, 6, 8, 10, 12}$,so $n(A) = 6$. Let $B$ be the event that the number is divisible by $3$. The numbers are ${3, 6, 9, 12}$,so $n(B) = 4$. The event $A \cap B$ is the set of numbers divisible by both $2$ and $3$ (i.e.,divisible by $6$). The numbers are ${6, 12}$,so $n(A \cap B) = 2$. Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{2}{12} = \frac{8}{12} = \frac{2}{3}$.

Twelve tickets are numbered $1$ to $12$. One ticket is drawn at random,then the probability of the number being divisible by $2$ or $3$ is:

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