In a class of 25 students with roll numbers 1 | vedclass

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(C) Let $A$ be the event that the roll number is a multiple of $5$,and $B$ be the event that the roll number is a multiple of $7$.The set of roll numb

Vedclass · Accepted Answer

$\frac{8}{25}$

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(C) Let $A$ be the event that the roll number is a multiple of $5$,and $B$ be the event that the roll number is a multiple of $7$.The set of roll numbers from $1$ to $25$ is $S = \{1, 2, 3, ..., 25\}$,so the total number of outcomes is $n(S) = 25$.The multiples of $5$ in the range are $A = \{5, 10, 15, 20, 25\}$,so $n(A) = 5$.The multiples of $7$ in the range are $B = \{7, 14, 21\}$,so $n(B) = 3$.Since there are no common multiples of $5$ and $7$ in the range $1$ to $25$,$A \cap B = \emptyset$,so $n(A \cap B) = 0$.The probability of the union of two events is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A) = \frac{5}{25}$,$P(B) = \frac{3}{25}$,and $P(A \cap B) = 0$.Therefore,the required probability is $P(A \cup B) = \frac{5}{25} + \frac{3}{25} - 0 = \frac{8}{25}$.

In a class of $25$ students with roll numbers $1$ to $25,$ a student is picked up at random to answer a question. Find the probability that the roll number of the selected student is either a multiple of $5$ or $7$.

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