geometric distrubtion and binomial confusion 
I'm confused with this question - here I'm assuming our discrete random variable is $X$ which is "the number of washing machine breakdowns in a year" is this correct? If so, what is $p = 0.8$ represent? Also, what is $P(X = k)$ I mean, is it the probability of getting $k$ breakdowns? If so - then I thought geometric distrubtion describes the probability of getting a success on your kth try - this doesn't seem to be the case here. A binomial seems more suitable. Why has the question chosen geometric
 A: The geometric distribution supported on the set $\{1,2,3,\ldots\}$ is the distribution of the number of independent trials needed to get one success with a fixed probability of success on each trial.  The number $p=0.8$ could be that probability of success on each trial, or it could be the probability of failure on each trial.  There is also a geometric distribution supported on the set $\{0,1,2,3,\ldots\}$ (this time including $0$) which is the distribution of the number of failures before the first success.
I know of no reason to regard each washing machine breakdown as a "trial" in a sequence of independent trials.  Thus it seems quite unnatural to use it to model the number of washing machine breakdowns.
A binomial distribution might be appropriate if, for example, one could potentially have just one breakdown per day (perhaps because only one visit by maintenance people could be done each day) or just one per month, etc.  The binomial distribution is the probability distribution of the number of successes in a fixed (i.e. non-random) number of trials.  If each day were such a trial and each breakdown a "success" then the binomial distribution could be used, with the number of days as the number of trials.  (But if they were actually independent of the history of breakdowns and repairs then perhaps the quality of the machines or maintenance work would be questionable, so that's one reason to think a binomial might not fit really well.)
So I see no reason why the geometric distribution makes sense.  But at least it is a distribution whose support is the set $\{0,1,2,3,\ldots\}$, and that is where the number of breakdowns would be.
PS: Alright, here's a fanciful, if not realistic, theory of why the geometric distribution supported on $\{0,1,2,3,\ldots\}$ would make sense.  Let's say some adjustable aspect of the machine has probability $p$ of being on the correct setting.  If it is correct, then the machine will run perfectly forever, but otherwise it will break down before the end of the year.  When it breaks down, the repair consists of randomly picking a setting, each time with probability $p$ of picking the right one.  The number of breakdowns could then be viewed as the number of failures before the first success.  That this is silly means that it is silly to use the geometric distribution to model something like this.  You'd do a bit better with the Poisson distribution, but that assumes the probability of a breakdown in the next ten minutes does not depend on how long it's been since the last breakdown.  If the waiting time between breakdowns had a gamma distribution that is the sum of several independent exponentially distributed random variables, that might not be too bad a model.
