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(A) Let the two positive integers be $x$ and $y$. Given $x + y = 100$,where $x, y \in \{1, 2, \dots, 99\}$.Total number of possible pairs $(x, y)$ is

Vedclass · Accepted Answer

$\frac{7}{9}$

Vedclass · Answer

(A) Let the two positive integers be $x$ and $y$. Given $x + y = 100$,where $x, y \in \{1, 2, \dots, 99\}$.Total number of possible pairs $(x, y)$ is $99$.We want the product $xy > 1000$.Substituting $y = 100 - x$,we get $x(100 - x) > 1000$,which simplifies to $100x - x^2 > 1000$,or $x^2 - 100x + 1000 Solving the quadratic equation $x^2 - 100x + 1000 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{100 \pm \sqrt{10000 - 4000}}{2} = \frac{100 \pm \sqrt{6000}}{2} = 50 \pm 10\sqrt{15}$.Since $\sqrt{15} \approx 3.87$,$x \approx 50 \pm 38.7$.So,$x$ must be in the range $(11.3, 88.7)$.The possible integer values for $x$ are $12, 13, \dots, 88$.The number of such integers is $88 - 12 + 1 = 77$.Therefore,the probability is $\frac{77}{99} = \frac{7}{9}$.

The sum of two positive integers is $100$. The probability that their product is greater than $1000$ is

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