Answer:
a) 0.42 = 42% probability that an alkaline battery will fail before 185 days
b) 0.12 = 12% probability that an alkaline battery will last beyond 2 years
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:

In which
is the decay parameter.
The probability that x is lower or equal to a is given by:

Which has the following solution:

In this problem, we have that:
. So


(a) What is the probability that an alkaline battery will fail before 185 days?


0.42 = 42% probability that an alkaline battery will fail before 185 days
(b) What is the probability that an alkaline battery will last beyond 2 years? (Use the fact that there are 365 days in one year.)
2 years = 2*365 = 730 days.
Either it lasts 730 or less days, or it last more. The sum of the probabilities of these events is decimal 1. So

We want P(X > 730). So

0.12 = 12% probability that an alkaline battery will last beyond 2 years