probabilitystatistics
Every day, a lecture may be cancelled due to inclement weather with probability 0.05. Class cancellations on different days are independent. Compute the probability that the tenth class this semester is the 3rd class cancelled?
This is a practice exam so the exact answer isn't near as important as the proper solution. My assumption is P(3rd cancelled is 10th class) = P(2 cancellations in 9 classes)*P(cancelled)
There is nice symmetry to exploit. Let $p$ be the probability of "more on Friday." Then $p$ is also the probability of more on Thursday. And $2p+e=1$, where $e$ is the probability the number of accidents is the same.
Once we know $e$, we will know $p$.
The probability the number of accidents is the same is the probability they are both $0$, plus the probability they are both $1$, plus the probability they are both $2$.
The probability they are both $0$ is $(0.6)^2$. Continue.
First, the problem as stated needs additional assumptions to have a clear-cut solution. Make sure you understand why, as it's quite key to understanding how to properly model a real-life problem with probability theory.
My personal take would be to see this as a latent variable / Hidden Markov Model (HMM) situation. General intro on HMM here : http://en.wikipedia.org/wiki/Hidden_Markov_model.
You could introduce a variable $Y_n$ corresponding to the $n$-th shot's outcome, and a hidden variable (or latent state) $X_n$ corresponding to the player's skill at time $n$. Assuming the player's success at time $n$ only depends on $X_n$ and not on $Y_{< n}$ (e.g. our guy does not stress out if he missed...), and $X_n$ only depends on $X_{n-1}$, our situation is perfectly described by the HMM below (illustration source here) : 
Now you still need to make a handful of modeling assumptions, namely :
Once you've done all this, your problem is one of inference, namely calculating $P(Y_n | y_1, ..., y_{n-1})$ (where we use the common convention: $Y$ is the random variable and $y$ is a realisation of $Y$). The method for resolving this is way beyond this site's scope of course ...
PS : That's just a very simple approach of HMM described in layman's terms. The roughest assumption made here is guessing/fixing the model's parameters once and for all. The bayesian way would be to leave all parameters (i.e. the $a_{m,k} = P(X_n = u^m | X_{n-1} = u^k)$ and $b_k = P(Y_n = \text{success} | X_n = u^k)$) open and then to calculate $P(Y_n | y_1, ..., y_{n-1})$ by averaging over the model's parameter space (given $y_1, ..., y_{n-1})$). This paper http://mlg.eng.cam.ac.uk/zoubin/papers/ijprai.pdf mentions this technique, look for the words "predictive distribution".
PPS : Another approach could be to see the situation as a mixture model (MM) problem. Wikipedia link : http://en.wikipedia.org/wiki/Mixture_model. It's far less appropriate than HMMs but you may find it easier to understand.
In a nutshell, the typical MM example is one of a bag of $K$ indistinguishable coins which are of two types, each type in an unknown proportion $a_1 K$ and $a_2 K$ and having a given unknown flip bias, say $b_1$ and $b_2$. You start to pick one coin at random, and you flip it $m$ times. Then you put the coin back in the bag and you start again, $n$ times. The goal is to determine best estimates of $\theta = (a_1, a_2 = 1-a_1, b_1, b_2)$ from the outcome of the experiment. Then, with your parameters estimated, it's easy to evaluate the distribution of your next marginal draw. The common technique to estimate $\theta$ is called Expectation maximization (EM) and is described quickly in the wikipedia link above. Intuitively, if you see in your data that in 20% (resp. 80%) of the draws, the proportion of faces is between 10% and 30% with an average of 22% (resp. between 55% and 70% with an average of 63%), you're tempted to conclude that $a_1 = 20\%$, $a_2 = 80\%$, $b_1 = 22\%$ and $b_2 = 63\%$. Also, if a draw has a majority of faces, it's highly probable that the coin in that draw was of type 2. EM formalizes that and does a bit more...
Now, you could say that in your problem the player is, for each game $n$, either Type 1 (good) or Type 2 (bad) (MM works with more than 2 categories but let's keep it at two), with the probability $p$ of him being good unknown to you, and that in each state his probability of success is $b_1$ and $b_2$. Run EM (or any other method), it will give you best estimates for $p$ and $b_i$. It will also give you, for each game $n$, the probability that the player was good in this particular game, say $q_n$. Here comes the fiddling : then plot $n \mapsto q_n$, hopefully it should be roughly increasing and come up with an extrapolation/approximation that will give you a guess (let's not call it an estimate) for $q_{n+1}$. A best guess for your probability of success in the future game $n+1$ is now $P(\text{success}_{n+1}) = q_{n+1} \hat b_1 + (1-q_{n+1}) \hat b_2$. Heck, it's really dirty, as some assumptions here are self-contradictory, but it's a start and my old physics teacher would have been satisfied with it.
Best Answer
The procedure you propose is correct.
Call a cancelled class a success. So we want the probability of exactly $2$ successes in $9$ trials, followed on the $10$-th trial by a success.
The probability of success is $p=0.05$. So the probability of exactly $2$ successes in $9$ trials is $\dbinom{9}{2}p^2(1-p)^7$.
Multiply by $p$ to get the desired probability.