heh, we are both wrong, but its a simple thing really.

Dantose, light bulbs are resistors, they just give off their power in terms of light, and we neglect heat in most cases.

Now, heres what happens:

In the first situation, you can minimise the resistor network (lightbulbs are resistors) to figure out what the current coming out of the battery is. You do this and you come up with a resistor of 4*R/3

Ohms law: V=I*R. V= I*4*R/3

Solving for I:

I= 3V/4R

That is the current across B1. Since the other 3 bubls are the same resistance, they get equal current, or V/4R.

Now if we remove one resistor from the parallel network, you get an equivilant resistor of 3R/2 By ohms law again:

I=2V/3R This is LESS Current than before, meaning that B1 gets dimmer.

But again, B2 and B4 are the same, meaning they get equal current, or I=V/3R, a LARGER amount of current than before, meaning they get brighter.

Calculating Brightness:

You can figure the brightness by the power dissipated. For a resistor, or a lightbulb, the easiest way to do this is P=I^2*R. In the first case:

B1: P=9V^2/16R

B2: P=V^2/16R

B3: P=V^2/16R

B4: P=V^2/16R

Without B2:

B1: P= 4V^2/9R

B2: P= V^2/9R

B4: P= V^2/9R

The easiest thing to do to get a hard number is go with a 1V batter, 1ohm resistor. For each of the bulbs above:

First case:

B1 = 9/16 Watt

B2 = B3 = B4 =1/16 Watt

Second Case:

B1 = 4/9 Watt

B2 = 1/9 Watt

So the end answer you want is D, that B1 gets dimmer, and B2 and B4 get brighter. Think of it this way, when you plug a vaccuum cleaner into the wall, you pull more current, and the other lights, that are in parallel, get dimmer.