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JEE Mains · Maths · STD 11 - 14. probability

Let \(M\) be the maximum value of the product of two positive integers when their sum is \(66\). Let the sample space \(S=\left\{x \in Z: x(66-x) \geq \frac{5}{9} M\right\}\) and the event \(A=\{ x \in S : x\) is a multiple of \(3\) \(\}\). Then \(P ( A )\) is equal to

  1. A \(\frac{15}{44}\)
  2. B \(\frac{1}{3}\)
  3. C \(\frac{1}{5}\)
  4. D \(\frac{7}{22}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{3}\)

Step-by-step Solution

Detailed explanation

\(M=33 \times 33\) \(x(66-x) \geq \frac{5}{9} \times 33 \times 33\) \(11 \leq x \leq 55\) \(A:\{12,15,18, \ldots .54\}\) \(P(A)=\frac{15}{45}=\frac{1}{3}\)
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