An integer is chosen at random from the integers 1,2,3, …, 50. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is

Solution

Maths sir · 3 steps

Count each set of multiples, then remove the overlaps using the LCM, not the product.

1. Count each set. Let A, B, C be the multiples of 4, 6 and 7 up to 50. Then n(A) = ⌊ 50/4 ⌋ = 12, n(B) = ⌊ 50/6 ⌋ = 8 and n(C) = ⌊ 50/7 ⌋ = 7.
2. Find the overlaps. A number divisible by both 4 and 6 is a multiple of LCM(4,6) = 12, so n(A ∩ B) = 4. Similarly LCM(4,7) = 28 gives n(A ∩ C) = 1 and LCM(6,7) = 42 gives n(B ∩ C) = 1. The LCM of all three is 84, beyond 50, so n(A ∩ B ∩ C) = 0.

   n(A ∪ B ∪ C) = (12 + 8 + 7) - (4 + 1 + 1) + 0 = 21

3. So. The probability is 21/50, which is option (d).

The idea to keep. For "at least one of", use inclusion-exclusion. Two numbers divide n together exactly when their LCM does, so 4 and 6 overlap at multiples of 12, not 24.

Answer from the NTA final answer key. Solution by PrepWiser.