I played with this and while the collector current is easy, the base currents need a little nodal analysis.

For the collector, assume both transistors are saturated and they drop V_{ce sat} of probably 0.2V.

Then you need to know something about the LED, specifically V_{f} (forward voltage drop) of, say 2.2V and you're just about there. The voltage across the resistor is 5 - 2*0.2 - 2.2 or 2.4V. Given a 1k resistor, there will be 2.4 mA flowing. That is probably nowhere near enough to turn on the LED.

Ordinary LEDs have a V_{f} of around 2.2V and an I_{f} of around 20 mA. That resistor would need to be 2.4V / 0.02A or about 120 Ohms. It could be a little larger with less LED current, perhaps as high as 220 Ohms.

Once you settle on the collector current, you can assume the transistors have an h_{FE} of 100 to get the minimum base current. Then you assume a V_{be} of 0.7V and you can do a nodal analysis on the resistors. In broad terms, the base current will need to be around 200 uA and still provide the appropriate base voltage.

If that is true, the lower resistor will drop about 2V. But it can't because the upper base voltage is 1.4V (two V_{be} drops). The most it can provide is 0.7V (one V_{be} drop) divided by 10k or 70 uA. Nowhere near enough if the collector current is to be 20 mA unless h_{FE} is a lot higher than 100.

A better solution is to put both resistors in parallel from the 5V rail. That series connection is a problem looking for a place to happen.

The other problem with the circuit is the floating base inputs. You might want to pull the bases down to ground with a 1 MOhm resistor. It could be smaller, you need to deal with it when you calculate the individual base resistors. Just remember, the upper base needs to get to 1.4V and the lower base needs to get to 0.7V and they need about 200 uA to turn on, given the assumptions.