in this video we are moving on to our third continuous probability distribution and we are talking about the exponential distribution so the exponential distribution models the time between independent events or a process time which is memoryless and so you can see that depending on our parameter Lambda which is our failure rate in units per hour that depending on our Lambda we get different shaped distributions the range for exponential distributions is always positive we cannot go less than zero we cannot have a negative value and our parameter of interest is Lambda and Lambda is equal to 1 over beta or vice versa and beta is our mean so that's our Mean Time between failure and so we have our probability function here we have 1 over beta time e raised to thex over beta and our mu is equal to Beta and our Sigma squar is equal to Beta squ so now we're going to hop into a problem based on extensive testing it is determined that the time Y in years before and major repairs required for a certain washing machine is characterized by the density function below note that Y is an exponential random variable with mu equal to four years so this four years means that the time between independent events so for this one the time between each failure of the washing machine is four years the machine is considered a bargain if it is likely unlikely to require a major repair before the sixth year so first question is what is the mean time between failur so that one's super easy that's our mu value and that's just going to be equal to four years then we want to know what is the probability of y greater than six what is the probability that we need a repair after that sixth year so for that we're going to find the integral because it's a continuous function we have to integrate and it wants to know greater than six so that's going to be everything from six to Infinity we're also going to stick with that Infinity here because our function is greater than or equal to zero so any number bigger than zero so we're going to go all the way up to Infinity in our integration and then that's going to be of f ofx DX so we're going to plug in our F ofx Formula so 1/4 e to the Y over 4 4 Dy so then we have to do some Uub here so we're going to have U equal to Y over 4 du is equal to -4 Dy which means Dy is equal to-4 du so then that gets us to -4 * 1/4 integral from 6 to Infinity of E U du e raised to the U so that gives us e to the U 6 Infinity which is equal toga e to the4 or negative y over four sorry evaluated from six to Infinity so now I can plug in my six and infinity values here so it's going to be equal to negative e to thetive Infinity minus E the6 over 4 so e raised to negative Infinity is going to take that to zero so then we're just going to end up with e -6 over 4 which is equal to 0.2 2 3 1 note that here when we are doing y greater than 6 we could also instead of doing six from Infinity we could also do 1 minus the integral from 0 to 6 that is an option here as well and so the probability of the first failure occurring after year six is 22% and so that also means that the probability of a failure occurring before year6 is 1 - 0.223 1 which is equal to 0.776 n so what that means is it's considered a bargain if it is unlikely to require a major repair before the sixth year so that bargain aspect applies to 22% which also means that 78% are not Bargains next question what is the probability that a major repair is required in the first year so that's asking what is the probability of 0 y1 so probability that we need a repair in that first year so this is going to be the integral from 0 to 1 of our function 1/4 e to the Y over 4 Dy so we would end up doing that same Uub as above and we end up getting Nega e to the Y over 4 evaluated from 0 to 1 which gives us e to the /4 minus E to the0 anything raised to zero goes to one so that's essentially equal to 1 minus E to the4 which is equal to 0.221 so we have a 22% chance up here at getting a bargain and then we also have a 22% chance of it requiring a major repair in the first year so the main takeaway here with exponential distribution is the application so we are specifically looking at the time between independent events and considering failure rate of whatever the context is so that's exponential distributions