Let the original electricity units used without energy savings be 100 units. After the first energy saving improvement, the energy used is 100 U - 25% of 100 U = 75U
After the second energy saving improvement, the energy used is 75U - 55% of 75 U = 33.75 U
After the third energy saving improvement, the energy used is 33.75 U - 20% of 33.75 U = 27U
Therefore, the overall percentage saving is 100U - 27U = 73U or 73%
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given bywhere Here we're given p=0.10 [ success = defective ] n=3