If a number is chosen at random from the set {x: 4 $\leq x \leq 15$}. Find the probability that it is a multiple of 3 or a multiple of 4

Analyze the question: We choose a number at random from the set {x : 4 ≤ x ≤ 15}. We need P(multiple of 3 or multiple of 4).

Concept: For "or" events, use P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Or simply list the favourable outcomes and count.

Solution:

Since x is meant to be an integer, the set is {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, so n(S) = 12.

• Multiples of 3: 6, 9, 12, 15 (4 numbers)
• Multiples of 4: 4, 8, 12 (3 numbers)
• Both (multiples of 12): 12 (1 number)

Numbers that are a multiple of 3 or 4: {4, 6, 8, 9, 12, 15}, which is 4 + 3 − 1 = 6 numbers.

P = 6/12 = 1/2

Answer: The probability is 1/2, which is option A.

Tip: In "or" probability questions, watch out for numbers counted twice (here, 12). Listing the favourable outcomes is a quick way to avoid double counting. Also, always count how many numbers are in the range carefully when both endpoints are included: 15 − 4 + 1 = 12.