The power supply is 6v.
Looking at the data sheet for that transistor, shows a VCE typ.~=90mV. @ IC of 10mA.
So lets use that data to try to analyse this circuit.
Since were using 10mA. for the collector current, then the voltage drop across the 470 ohm resistor will be 4.7V. Also from the Data sheet for the transistor, we can assume a value of 90mV. for the drop across this transistor, when it is saturated.
That makes a toal of ~=4.8v.
Subtract that from the power supply voltage and we have ~=1.2V. left over which is the drop across the flashing LED.
Now to make this transistor produce this current, we have to introduce a current that is around 20 times less than it's collector current. This is the base current.
The data sheet shows a value of IB to be 0.5mA. when IC is 10mA.
Now what value of resistance would be needed across the probes to introduce 0.5mA. of current into the base.
The data sheet shows 700mV. for the VBE, under saturated conditions. (transistor is used as a switch).
So to calculate the resistance needed across the probes would be like this.
(VCC - VBE) / IB = Rprobe. (6V. - 0.7V.) / 0.5mA. = 10600 ohms.
Now follow carefully what this is showing.
If you put a resistance of around 10 thousand ohms across the probes, it will allow a very small current of 0.5mA. to flow through these probes. BUT this small current through these probes, is also flowing into the base of the transistor to turn it on. When this transistor is turned on by this amount of current into its input base, then it opens up a vast amount of current iup to 10mA. to flow from its collector to emitter, since this collector to emitter of the transistor is in series with the LED, and the LED must have at least 10mA. for it to work, than this 10mA. that the transistor is supplying is able to work the LED.
WHILE at the same time it only took a current of 0.5mA to make all this happen.
Remember this 0.5mA. can NOT turn on the LED, it is too small of a current.
So why the transistor in the circuit, it acts like a relay, its base circuit is in effect a seperate circuit like a relay coil, ant the collector to emitter is like the relay contacts to the load being driven.
So it is like a relay, only the terminals are all connected with the same voltage, however the base loop is seperate from the collector emitter loop, as it is a branch off of the collector emitter loop.
The emitter always has the full amount of current flowing through it, while a small portion flows into the base, and the rest into the collector.
So see if that helps better understand the purpose of the transistor in this circuit.
PS,:
as adioguru said the 47K won't allow saturation.
I forgot to add the 47K resistor into the equation.
That should be 4.7K ohms so that a 5K resistance across the probes would make this circuit work.
But you get the idea of how thios could work with the proper resistor values.