Can they pick the same number??

Each of the numbers $1$ to $100$ has a probability of $\frac{1}{100}$ of being picked by Radha. If Radha picks ball $n$ then the probability that Shyam draws a ball with a larger number is

$\frac{100 - n}{100} = 1 - \frac{n}{100}$ .

Thus the probability that Shyam draws a ball with a larger number than Radha's will be

$x = (\frac{1}{100})*\sum_{n=1}^{100} (1 - \frac{n}{100}) =$

$(\frac{1}{100})*(100 - (\frac{1}{100})*\sum_{n=1}^{100} n) =$

$1 - (\frac{1}{100*100})*(\frac{100*101}{2}) = 1 - \frac{101}{200} = \frac{99}{200}$ .

Thus $1000x = 1000*(\frac{99}{200}) = \boxed{495}$ .

Alternately, note that the probability that Shyam picks the same ball as Radha is $\frac{1}{100}$ , and thus the probability that they pick different balls is $\frac{99}{100}$ . Since the probabilities will be the same if Shyam picked first and Radha second, the probability that Shyam picks a ball with a larger number is just half the probability that they pick different numbers, i.e., $(\frac{1}{2})(\frac{99}{100}) = \frac{99}{200}$ , as found before.

The probability to pick the first ball is $\frac{1}{100}$ .

If we picked ball1, then there are 99 balls bigger than ball1.

So, the probability was $\frac{1}{100} * \frac{99}{100}$ .

If we picked ball2, then there are 98 balls bigger than ball2.

So, the probability was $\frac{1}{100} * \frac{98}{100}$

If we picked ball3, then there are 97 balls bigger than ball3.

So, the probability was $\frac{1}{100} * \frac{97}{100}$

We can see that if we picked ball $n$ , then there are 100 - $n$ balls bigger.

So, the probability must be:

$\frac{1}{100} * \frac{99}{100} + \frac{1}{100} * \frac{98}{100} + \frac{1}{100} * \frac{97}{100} + \frac{1}{100} * \frac{96}{100} + \ldots + \frac{1}{100} * \frac{2}{100} +\frac{1}{100} * \frac{1}{100}$

$= \frac{1}{100} (\frac{99 + 98 +97 +96 +95 + \ldots + 3 +2 +1}{100})$

And $1+2+3+\ldots+97+98+99 = \frac{99(100)}{2} = 4950$

And we get $x = \frac{1}{100} * \frac{4950}{100} \Rightarrow 1000x = \boxed{495}$

If they pick the same number, Shyam cant draw a bigger ball. This happens with a chance of $\frac{1}{100}$ . Thus the probability of them choosing different numbers is $\frac{99}{100}$ . Note that now we can imagine choosing two random balls, and then assigning one to Shyam and one to Radha. Obviously there is a $\frac{1}2$ chance that Shyam will get the bigger ball, so thus the total probability is $\frac{99}{100}\cdot\frac{1}2=\frac{99}{200}$ and the answer is $1000\left(\frac{99}{200}\right)=\boxed{495}$

If Radha picks 1, then Shyam has to pick any ball numbered 2 to 100. i.e., 99 possibilities. similarly if she picks 2, then he has to pick any number between 3 to 100 i.e, 98 possibilities. This goes on till she picks 99 numbered ball after which he has to pick 100th ball i.e,. a single possibility. So, total number of possible ways are 99+98+97+. . . . . .+1 which is the Summation of natural numbers from 1 to 99. That gives us 4950. This has to be divided by Total number of all events i.e, 100 x 100 (cos, She puts the ball back). thus x=4950/(100 x 100)=0.495 and 1000x=495

Radha can pick one $1$ to $100$ , and Shyam can pick $1$ to $100$ then total order pairs is $100*100$ . $(E)=100000$ The number of cases where Shyam draws a ball that has a bigger number than Radha's are equal to the number of cases where Radha draws a ball that has a bigger number than Shyam's, and exist $100$ order pairs which both numbers are equal. So, total of cases which Radha draws a ball that has a bigger number than Shyam's can be $\frac{100000-100}{2}=4950$ Finally, requiered probility is $\frac{4950}{1000}=0.495$ and $1000x=\boxed{495}$

This problem also can be solved using geometry (areas)