Question

n electric vehicle has a battery pack that is made up of 10 individual cells. With a 95% probability, we know that any given cell will last at for least 10,000 charge-discharge cycles. Assuming that the failure of any given cell in the battery pack does not depend, in any way, on the performance of any other cell in the battery pack, what is the probability that every cell in the battery pack will last for at least 10,000 charge-discharge cycles

Answer 1

0.5987 = 59.87% probability that every cell in the battery pack will last for at least 10,000 charge-discharge cycles

Explanation:

For each cell, there are only two possible outcomes. Either it will least for at least 10,000 charge-discharge cycles, or it will not. Cells are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

10 individual cells.

This means that
`n = 10`

95% probability

This means that
`p = 0.95`

What is the probability that every cell in the battery pack will last for at least 10,000 charge-discharge cycles

This is P(X = 10).

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 10) = C_(10,10).(0.95)^(10).(0.05)^(0) = 0.5987`

0.5987 = 59.87% probability that every cell in the battery pack will last for at least 10,000 charge-discharge cycles