Electrical – A x10 oscilloscope probe means that the voltage is attenuated 10 times

One simple example

Let's say you have two scope probes. Probe A is correctly grounded to the circuit you are measuring but probe B doesn't have a ground connection. Let's also say that probe A is measuring a 1Vp-p squarewave on the circuit and probe B is looking at a DC level.

Despite scopes having a 1M\$\Omega\$ input impedance, there will be currents that flow through the probe due to the edges of the square wave - the scope input impedance is also likely to comprise 20pF capacitance (smaller for x10 probes usually).

This capacitance "takes" a pulse of current down the probe lead every time the square wave alternates causing a small volt-drop down the lead. This won't matter for probe A but, this will appear as a small "spike" on channel B's display because the glitch "becomes" in series with the probe B signal due to it being ungrounded.

If probe B were correctly grounded at the circuit this "glitch" won't be present on the scope B waveform.

What point/frequency does this become significant? I would answer that by saying just don't take short cuts. If you are probing a circuit for whatever reason, use best practise and ground both probe ends at the circuit you are measuring or expect anomalies.

If probe A is looking at 1Vp-p squarewave and probe B is looking at a 5mV analogue signal you'll have troubles on channel B.

Ignore the capacitors, cable etc. and just consider the resistors. From probe tip to ground you have 9MΩ in series with 1MΩ - that's 10MΩ! You can verify this by measuring the resistance between probe tip and ground with a multimeter (on 20MΩ scale).

Of course this only applies to DC. At higher frequencies the impedances of the capacitors and cable become significant. At 50MHz the reactance of a 10pF capacitor is about 300Ω.