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 event of getting a multiple of $5$. The multiples of $5$ between $1$ and $30$ are ${5, 10, 15, 20, 25,

Vedclass · Accepted Answer

$\frac{1}{3}$

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(B) Total number of balls $= 30$.Let $A$ be the event of getting a multiple of $5$. The multiples of $5$ between $1$ and $30$ are ${5, 10, 15, 20, 25, 30}$. So,$n(A) = 6$.Let $B$ be the event of getting a multiple of $7$. The multiples of $7$ between $1$ and $30$ are ${7, 14, 21, 28}$. So,$n(B) = 4$.Since there are no common multiples of $5$ and $7$ in the range $1$ to $30$,$n(A \cap B) = 0$.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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