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14.Probability
hard
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
A
$\frac{15}{44}$
B
$\frac{1}{3}$
C
$\frac{1}{5}$
D
$\frac{7}{22}$
(JEE MAIN-2023)
Solution
$M=33 \times 33$
$x(66-x) \geq \frac{5}{9} \times 33 \times 33$
$11 \leq x \leq 55$
$A:\{12,15,18, \ldots .54\}$
$P(A)=\frac{15}{45}=\frac{1}{3}$
Standard 11
Mathematics
