There are two light bulbs in a circuit wired in parallel a third light bulb is added again in parallel what is the effect on total resistance of adding the third bulb?

Total resistance decreases:

1/R(total) = 1/R1 + 1/R2 + 1/R3

Assuming each lightbulb has the same resistance: R1 = R2 = R3

1/R(total) = 1/R = 1/R + 1/R = 3/R

R(total) = R/3

Before the bulb was added:

1/R(total) = 1/R + 1/R = 2/R

R(total) = R/2

R/3 < R/2