@AamirshahSHAIK due to how most digital inputs are fabricated, they cannot be left floating. This is because the input is essentially a very high-gain amplifier; what happens when you increase a radio volume to max with nothing playing? You hear static - this causes voltage transients in the digital circuitry which can lead to oscillation. Also, digital inputs consume far more current when inbetween the voltages it sees as "low" and "high."

To see this for yourself, place a 4000-series CMOS chip into a breadboard and tie all of it's inputs to ground through 10k resistors. Measure the power consumption and if possible, probe it with an oscilloscope. Then remove an input pull-down and scope again. The chip may be destroyed, careful.

but why does it go away when i connect the pin to ground , there are no electrons free arround that can just go to ground from the pin

i understand apin that is not connected to gnd or +V is affected by electrical noise , but what is the physical thing that magically once you connect to gnd or+V it's not affected by electrical noise

The "magical thing" is input impedance. This means, the amount of resistance the input has. For CMOS and other digital inputs, the input impedance is very high - megaohms or more - meaning that any tiny thing (even stray electrons) can influence it. Which is why it is so sensitive to noise. But tie a 10k pull-down to it, and the noise and stray electrons are much smaller than the pull-down, so never cause the input to leave "low" status.

Another analogy might be that of a small gasoline (petrol) engine. A weak spring (pull-down) is used to normally keep the throttle at idle speed. Increase the throttle (against the weak spring), and the engine accelerates. So what happens if this spring is removed? It only takes a microgram to turn the throttle valve - the engine may accelerate to any speed or even oscillate because the valve is SO sensitive.