Question

If a number is chosen from the first 150 natural numbers, then find the probability that the number is either a multiple of \(3\) or \(4\).

• A. \(\dfrac{1}{3}\)
• B. \(\dfrac{1}{4}\)
• C. \(\dfrac{1}{2}\)
• D. \(\dfrac{1}{5}\)

Hint:

For probability involving “or”: \[ \boxed{n(A\cup B)=n(A)+n(B)-n(A\cap B)} \] Remember to subtract the common multiples once to avoid double counting.

Answer 1

The Correct Option is C

Concept: Use the principle of inclusion--exclusion. \[ \boxed{n(A\cup B)=n(A)+n(B)-n(A\cap B)} \] where
• \(A\): Multiples of \(3\)
• \(B\): Multiples of \(4\)

Step 1: Find the multiples of \(3\).
\[ \left\lfloor\frac{150}{3}\right\rfloor=50. \] So, there are \(50\) multiples of \(3\).

Step 2: Find the multiples of \(4\).
\[ \left\lfloor\frac{150}{4}\right\rfloor=37. \] So, there are \(37\) multiples of \(4\).

Step 3: Find the common multiples.
The LCM of \(3\) and \(4\) is \[ 12. \] Hence, \[ \left\lfloor\frac{150}{12}\right\rfloor=12. \] Thus, there are \(12\) numbers that are multiples of both \(3\) and \(4\).

Step 4: Apply inclusion--exclusion.
\[ n(A\cup B) =50+37-12 =75. \] Therefore, \[ \text{Probability} =\frac{75}{150} =\frac12. \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac12} \] is the required probability.