Natural selection!

Let`s see that $x+\frac { 100 }{ x } >50\Longrightarrow { x }^{ 2 }+100>50x$ now we can rewrite the equation as $x^{ 2 }-50x>-100\Longrightarrow x(x-50)>-100$ .

Now, is important to know that the first $100$ natural numbers are $1,2,3,...,100$ .

See, that the values that satisfy the equation $x(x-50)>-100$ are $x\ge 48$ and $x=1,2$ .

So, we have a total of $55$ solutions in a total of $100$ numbers; so the probability is $\frac { 55 }{ 100 } =\frac { 11 }{ 20 } =\frac { a }{ b }$ so, clearly, $a+b=31$ .

Your question is correct but your answer is wrong.... First hundred natural numbers are 0 to 99.....in that case our expected numbers belong to {1,2 and 48 to 99}.... Anyways I did it assuming you will consider it as 1 to 100

x+100/x>50 :::> x^2-50x+100>0 since the quadratic equation x^2-50x+100=0 has two roots 2.087 and 47.91 and the coefficient of x^2 is positive.Therefore it will satisfy the condition only when x is less than2.087 or greater than 47.91 .And the first natural numbers which satisfy it are {1,2,48,49,50,51,51,.........,99,100}. Therefore probability will be 55/100=11/20. Therefore a+b=11+20=31.

Without using the Python sympy library I would have made the mistake of missing the contributions of probabilistic mass for the low values of x .

>>> solve univariate inequality(x+100./x>50,x)

Or(x <= 2.0871215252208, x > 47.9128784747792)

There are 55 x-values in this set and each contributes 0.01 of probability. Hence the result.

Since I can never trust my manipulations I also did a quick Monte-Carlo as a check:

>>> count = 0

>>> for i in range ( 1000000 ) :

... r = randint ( 1, 100 )

... count += ( ( r + 100. / r ) > 50 )

...

>>> count

549730

>>> 549730./1000000

0.54973

Pretty close agreement.

$P=Pr(x+\frac {100} x > 50; x\in\{1,\,2,\,...,100\}) = \frac {N_{OK}}{N_{TOT}}=\frac {N_{OK}} {100};\,$

$x+\frac {100} x > 50,\,\Rightarrow x^2-50x+100 > 0;\,$

$x_{1,2}=25\pm\sqrt{525},\,x_1\approx 2.087,\,x_2\approx 47.913$

The set of solutions is:

$\{x < x_1\}\cup\{x > x_2\}\cup\{x\in\{1,\,2,\,...,100\}\},$

$\{x\le 2\}\cup\{x\ge {48}\}\cup\{x\in\{1,\,2,\,...,100\}\},$

$x\in\{1,\,2\}\cup\{48,\,49,\,...,100\},$

$N_{OK}=2+53=55$

$P=\frac {55}{100} = \frac {11}{20} = \frac a b$

$a+b=11+20=\boxed{31}$

The first 100 natural numbers are 1-100.

Starting from 1, we can substitute each value in the inequality. (Trust me, won't take long to find a pattern)

1 + 100/1 = 101

2 + 100/2 = 52

3 + 100/3 = 36.33...

For values 3 and after, the inequality does not hold. Until, obviously x has a value nearby 50. For 50 and all values 50-100, the inequality will obviously hold.

49 + 100/49 = 49 + 2.xyz = 51.xyz

48 + 100/48 = 48 + 2.jkl = 50.jkl

47 + 100/47 = 47 + 2.abc = 49.abc

Thus, for 45 values from 3-47, the inequality doesn't hold.

The remaining probability is 55/100 = 11/20 a + b = 31.