If a number is selected from the first 30 nat | vedclass

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(D) The set of the first $30$ natural numbers is $S = \{1, 2, 3, \dots, 30\}$,so the total number of outcomes is $n(S) = 30$. Let $A$ be the event tha

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) The set of the first $30$ natural numbers is $S = \{1, 2, 3, \dots, 30\}$,so the total number of outcomes is $n(S) = 30$. Let $A$ be the event that the number is divisible by $4$. The numbers are $\{4, 8, 12, 16, 20, 24, 28\}$,so $n(A) = 7$. Let $B$ be the event that the number is divisible by $7$. The numbers are $\{7, 14, 21, 28\}$,so $n(B) = 4$. The event $A \cap B$ represents numbers divisible by both $4$ and $7$ (i.e.,divisible by $28$). The only number is $\{28\}$,so $n(A \cap B) = 1$. Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $P(A \cup B) = \frac{7}{30} + \frac{4}{30} - \frac{1}{30} = \frac{10}{30} = \frac{1}{3}$.

$P(A)$	$P(B)$	$P(A \cap B)$	$P(A \cup B)$
$0.5$	$0.35$	$.........$	$0.7$

If a number is selected from the first $30$ natural numbers,then the probability that the number selected is divisible by $4$ or $7$ is:

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