A box contains 36 tickets numbered 1 to 36. O | vedclass

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(B) Let $A$ be the event that the number is divisible by $3$.$A = \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36\}$,so $n(A) = 12$.Let $B$ be the event

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$\frac{4}{9}$

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(B) Let $A$ be the event that the number is divisible by $3$.$A = \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36\}$,so $n(A) = 12$.Let $B$ be the event that the number is a perfect square.$B = \{1, 4, 9, 16, 25, 36\}$,so $n(B) = 6$.The intersection $A \cap B$ contains numbers divisible by $3$ and are perfect squares.$A \cap B = \{9, 36\}$,so $n(A \cap B) = 2$.The total number of outcomes is $36$.Using the addition rule for probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = \frac{12}{36} + \frac{6}{36} - \frac{2}{36} = \frac{16}{36} = \frac{4}{9}$.

$A$ box contains $36$ tickets numbered $1$ to $36$. One ticket is drawn at random. Find the probability that the number on the ticket is either divisible by $3$ or is a perfect square.

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