A bag contains 30 balls numbered from 1 to 30 | vedclass

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(B) Total number of balls $= 30$.Let $A$ be the set of numbers that are multiples of $5$ between $1$ and $30$:$A = \{5, 10, 15, 20, 25, 30\}$,so $n(A)

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$\frac{1}{3}$

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(B) Total number of balls $= 30$.Let $A$ be the set of numbers that are multiples of $5$ between $1$ and $30$:$A = \{5, 10, 15, 20, 25, 30\}$,so $n(A) = 6$.Let $B$ be the set of numbers that are multiples of $7$ between $1$ and $30$:$B = \{7, 14, 21, 28\}$,so $n(B) = 4$.The intersection $A \cap B$ contains numbers that are multiples of both $5$ and $7$ (i.e.,multiples of $35$):$A \cap B = \emptyset$,so $n(A \cap B) = 0$.The number of favorable outcomes is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 6 + 4 - 0 = 10$.The probability $P(A \cup B) = \frac{n(A \cup B)}{	ext{Total outcomes}} = \frac{10}{30} = \frac{1}{3}$.

$A$ bag contains $30$ balls numbered from $1$ to $30$. One ball is drawn randomly. The probability that the number on the ball is a multiple of $5$ or $7$ is:

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