Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = \left\{ {x \in \mathbb{Z}:x(66 - x) \ge {5 \over 9}M} \right\}$ and the event $\mathrm{A = \{ x \in S:x\,is\,a\,multiple\,of\,3\}}$. Then P(A) is equal to :
Solution
$x+y=66$
<br/><br/>
$$
\begin{aligned}
& \frac{x+y}{2} \geq \sqrt{x y} \\\\
\Rightarrow & 33 \geq \sqrt{x y} \\\\
\Rightarrow & x y \leq 1089 \\\\
\therefore & M=1089 \\\\
S: & x(66-x) \geq \frac{5}{9} \cdot 1089 \\\\
& 66 x-x^{2} \geq 605 \\\\
\Rightarrow & x^{2}-66 x+605 \leq 0
\end{aligned}
$$
<br/><br/>$\Rightarrow(x-55)(x-11) \leq 0 ; 11 \leq x \leq 55$
<br/><br/>Therefore $S=\{11,12,13 \ldots 55\} $
<br/><br/>$\Rightarrow n(S)=45$
<br/><br/>Elements of $S$ which are multiple of 3 are
<br/><br/>$$
\begin{aligned}
& 12+(n-1) 3=54 \Rightarrow 3(n-1)=42 \Rightarrow n=15 \\\\
& n(A)=15
\end{aligned}
$$
<br/><br/>$\Rightarrow P(A)=\frac{15}{45}=\frac{1}{3}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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