Question

The battery cells are heavy and if they do not work, they are no more than an extra load that puts the functionality of electric cars at risk. So, if the liability

of cells is low, adding the number of cells can be a bad idea. We summarize this fact to this rule that for an electric car to be functional, the majority of

cells must work. If each cell breaks with the probability of ?, for which value of ?, a car with 5 battery cells is a better option than a car with 3 cells?

Answer 1

For car with 3 cells, number of broken cells X ~ Bin(n = 3, q)

probability that majority of cells must work = P(X 1) = P(X = 0) + P(X = 1)

For car with 5 cells, number of broken cells X ~ Bin(n = 5, q)

probability that majority of cells must work = P(X 2) = P(X = 0) + P(X = 1) + P(X = 2)

If car with 5 battery cells is a better option than a car with 3 cells,

probability that majority of cells must work for 5 battery cells > probability that majority of cells must work for 3 cells

which is true for,

q > 0 and 1-2q > 0 (Note that q is probability and cannot be less than 0)

q > 0 and q < 1/2

0 < q < 1/2