How to determine which bulb will be the brightest in the series and parallel connection? I stumbled upon these sentences while studying and this seems to be horribly confusing.

resistance is proportional to the inverse of the power of the bulb in series connection and so the bulb with the lowest wattage(power) will have maximum resistance and it will glow the brightest.

But what i don't seem to understand is that the brightness of the bulb should depend on the power or heat dissipated(in case of an incandescent light bulb) then why is it that in this case the bulb with the lowest power or the highest resistance glows the brightest.
Moreover in the next paragraph, which is for bulbs in parallel connection, it is said that

As the resistance of the highest wattage (power) bulb is minimum, it will glow the brightest.

Another question that arises at this point is that why are the two paragraphs contradicting?
From both the paragraphs it is clear that the relationship used is 

resistance is proportional to the inverse of the power  $$P ∝ \frac{1}{R}$$ 

but then why wasn't the other two formulas used

$$P= VI = {I^2}R$$

according to

$$P= {I^2}R$$

resistance is directly proportional to the electric power so shouldn't the power increase with increase in resistance or vice-versa.
One thing that is to be kept in mind is that the bulbs were manufactured for working of the same voltage.
 A: When the book says:

As the resistance of the highest wattage (power) bulb is minimum, it
  will glow the brightest.

What it is referring to is the power rating of the bulb.  This power rating assumes the bulb has a potential difference across it of 120 V (in North America at least - maybe 220 V in India).  So, if you compare a bulb rated at 60 watts and a bulb rated at 100 watts the 100 watt bulb will have a lower resistance and draw more current when they are connected in parallel (they way they are designed to used).
If, on the other hand, you connect the bulbs in series, then what is constant is the current not the voltage.  The bulb with the highest power rating, which has the lowest resistance, will thus have the smaller voltage drop across it and thus will dissipate the smaller amount of power.
A: 
. . . . the bulb with the lowest wattage(power) will have maximum
  resistance and it will glow the brightest.

$R= \dfrac {V^2}{P}$ so for a given supply voltage $V$ the bulb with the higher power rating will have a lower resistance.  
When two bulbs are connected in series to a power supply, the current $I$ through both bulbs is the same.
As the power dissipated in a bulb is $I^2\,R$ the bulb with the higher resistance (lower power when connected directly across the supply) will dissipate the greater power.  

The analysis above assumes that the resistance of the bulbs does not depend on the voltage across them.
This is in fact not true and if the voltage does not change by too much, the variation of the resistance of a tungsten filament bulb is approximately $R \propto V^{0.5}$ and the power dissipated $P \propto V^{1.5}$.  
There is more about the properties of such a bulb in the answer to the question entitled Car headlight voltage.
