Letters and envelopes probability (a) If three letters are placed at random in three envelopes, what is the probability that exactly one letter will be placed in the correct envelope? 
(b) If n letters are placed at random in n envelopes, what is the probability that exactly n−1 letters will be placed in the correct envelopes
For (a) I was thinking it will be $${3 \choose 1}\cdot(1)^{1/3}\cdot(2)^{2/3}$$
is that right? what would it be for a general n term then?
 A: For (a), split it into disjoint events and then add up their probabilities:


*

*The probability that only the 1st letter is in the correct envelope is $\frac16$

*The probability that only the 2nd letter is in the correct envelope is $\frac16$

*The probability that only the 3rd letter is in the correct envelope is $\frac16$


The overall probability is therefore:
$$\frac16+\frac16+\frac16=\frac12$$

For (b), the probability is obviously $0$, since if $n-1$ letters are in the correct envelope, then the remaining letter has "nowhere else to go" but the correct envelope too...
A: For your version of (a) with $n = 3,$ simply consider the six 
possible arrangements: 123, 132, 213, 231, 312, 321. 
 Of these 132, 321, and 213 have exactly one letter in
the correct position (letters 1, 2, and 3, respectively).
(b) Never possible to have exactly $n - 1$ in in proper
envelopes. One letter in improper envelope implies another
must also be improperly placed.  
A: If we consider a function y=x ,then basically from this point of veiw we can say that we have to match its input with output ,say we have input a,b,c then our output will be a,b,c respectively .Therefore what we have to to do is just mismatch 2 input with it's output i.e one subset is (a,a),(b,c),(c,b) and there will be similar two other subset .And total ways of doing this will be 3! so Our probability will be 3/6=1/2
For second part of the question it's probability is 0. As it is not possible to put 1 letter in incorrect postion as it will take correct position of other .
