random variables involving students doing a test A student answers a set of 100 true-false questions by answering 36 questions correctly and guessing the other
64. Another student also knows 36 correct answers and guesses the rest at random. What is the probability
that only one of the two students passes?
Attempt of solving it:
the probability of getting a right answer is (1/2) and the probability of getting a wrong answer is (1/2)
 so does P(R) = (1/2)^36
but I am not quite sure how to proceed from this 
How could you find out which student passes?
Any hints would be highly appreciated
Thanks!!
 A: The two students, say A and B, have the same probability of passing. Call that probability $p$. Only one of them passing can happen in two ways: (i) A passes and B fails or (ii) the other way around. 
The probability of (i) is $p(1-p)$, as is the probability of (ii). So the probability exactly one passes is $2p(1-p)$.  
Now let's find $p$, or at least start finding $p$. Student A passes if she gets the right answer on  $14$ or more of the $64$ questions where she guesses. That's all she needs, since she already has a sure $36$. 
It is a little easier to get at $1-p$, the probability she gets $13$ or fewer questions right. For that you need to know something about the Binomial distribution. The probability she gets $k$ of the $64$ questions right is the same as the probability of getting exactly $k$ heads in $64$ tosses of a fair coin. This is $\binom{64}{k}\left(\frac{1}{2}\right)^{64}$. We conclude that 
$$1-p=\sum_{k=0}^{13}\binom{64}{k}\left(\frac{1}{2}\right)^{64}.$$
Not too bad to evaluate, using appropriate software. But it may be that you are intended to use the normal approximation to the binomial. Note that if $X$ is the number of questions A guesses correctly, then $X$ has mean $32$ and standard deviation $\sqrt{(64)(1/2)(1/2)}=4$.
In terms of standard deviation units, $13$ or less is awfully far from the mean $32$. So for all  practical purposes, the probability of failure is $0$.  
Remark We did a fair bit of detail to give some information about situations when things are not so clear cut.  
The answer is intuitively obvious: both are essentially sure to pass, only one person passing is extremely unlikely, Failure happens only if we get $13$ or fewer heads in $64$ tosses of a fair coin. That's very highly unlikely. The teacher is probably quite popular.  
A: You didn't specify the criteria for passing, but let's say it's answering 50 out of 100 questions correctly. So each student essentially knows 36 out of 100, so all they need is to guess any 14 or more out of remaining 64. The probability that student A or B (they are equal) pass is 
$$
P(A)=P(B)=\sum_{k=14}^{64}\binom{64}{k}0.5^{k}0.5^{64-k}=0.5^{64}\sum_{k=14}^{64}\binom{64}{k}
$$
which does not exist in closed form (this is called partial sum of rows of Pascal triangle). Assuming students work independently, what you need is the probability that either of them passes, which is 
$$
P(A \cup B)=P(A)+P(B)-P(A)P(B)=2P(A)-(P(A))^2
$$ 
Can you handle from here?
