Probability that you have exactly one correct answer on a test I have two questions from a practice test which I have some concerns about:

Assume you write a multiple-choice exam that consists of 100 questions. For each question,
  4 options are given, one of which is the correct one. If you answer each of the 100 questions
  by choosing an answer uniformly at random, what is the probability that you have exactly
  one correct answer?
(a) $\frac{100}{4^{100}}$
  (b) $\frac{3^{99}}{4^{100}}$
  (c) $\frac{100+3^{99}}{4^{100}}$
(d) $\frac{100\cdot3^{99}}{4^{100}}$

The answer is (d). Can someone help me understand how to go about this problem?  


*

*How does  $100\cdot3^{99}$ count the number of ways there can be exactly one correct answer?

*I know that $3^{99}$ is the number of ways to choose $3$ possible answers from $99$ questions, while $1$ question is already fixed (correct). Why multiply by $100$?




You flip a fair coin 5 times. Define the events  
$A =$ "the number of heads is odd"  
and   
$B =$ "the number of tails is even"
(a) $Pr(A) = Pr(B)$
  (b) $Pr(A) < Pr(B)$
  (c) $Pr(A) > Pr(B)$  

The answer is (a). 


*

*My initial understanding was that, the numbers $1$ to $5$ consist of $1, 2, 3, 4, 5$. There are $3$ odd numbers and $2$ even numbers. So $Pr(A) > Pr(B)$.


For (a): since the answer is selected uniformly at random, the probability of $one$ correct answer out of 4 is $p=\frac{1}{4}, q=1-p= \frac{3}{4}$. Now, you have to select $exactly$ one out of 100, there are  $\binom{100}{1}$ ways to do it. 
Now, to get the probability of exactly one correct answer out 100 with this setup is $\binom{100}{1}pq^{99}$, hence the solution.  
