Stats binomial probability holiday lights

Question

A holiday light string has 27 lights, and a whole string fails if and only if one or more of its own individual lights fail. Each single light has an independent 2.7% chance of failure during any one year period. You buy two such light strings (one with all red lights, the other all green), and will simultaneously power each string for a full year. At year end, what is the percentage probability that the red string will be bright, and the green string will have failed?

Answer 1

Use the binomial distribution formula:

Pr ( r successes in n trials ) = number of combinations of r objects from n objects x the probability of r successes x the probability of ( n- r ) failures. Here we define success as light bulb does not fail.

Pr ( red string does not fail ) = number of combinations of 0 items from 27 items x .027⁰ x ( 1-.027 )²⁷. =

1 x 1 x .973²⁷ = .4775

The Pr ( green string fsils ) = 1 - Pr ( green string does not fail ) = 1- .4775 = .5225

So since the red string and green string outcomes are independent events, we multiply their probabilities.

Pr ( red string does not fail and green string fails ) = .4775 x .5225 = .2494 or almost 25%.