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(B) Let $S = \{1, 2, 3, \ldots, 50\}$,so the total number of outcomes is $n(S) = 50$.Let $A, B,$ and $C$ be the sets of multiples of $4, 6,$ and $7$ i

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$\frac{21}{50}$

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(B) Let $S = \{1, 2, 3, \ldots, 50\}$,so the total number of outcomes is $n(S) = 50$.Let $A, B,$ and $C$ be the sets of multiples of $4, 6,$ and $7$ in $S$ respectively.$A = \{4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48\} \implies n(A) = 12$.$B = \{6, 12, 18, 24, 30, 36, 42, 48\} \implies n(B) = 8$.$C = \{7, 14, 21, 28, 35, 42, 49\} \implies n(C) = 7$.Now,find the intersections:$A \cap B = \{12, 24, 36, 48\} \implies n(A \cap B) = 4$.$B \cap C = \{42\} \implies n(B \cap C) = 1$.$A \cap C = \{28\} \implies n(A \cap C) = 1$.$A \cap B \cap C = \emptyset \implies n(A \cap B \cap C) = 0$.Using the inclusion-exclusion principle:$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$.$n(A \cup B \cup C) = 12 + 8 + 7 - 4 - 1 - 1 + 0 = 21$.The required probability is $\frac{n(A \cup B \cup C)}{n(S)} = \frac{21}{50}$.

An integer is chosen at random from the integers $\{1, 2, 3, \ldots, 50\}$. The probability that the chosen integer is a multiple of at least one of $4, 6,$ and $7$ is

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