# 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?

• How do you know which is the correct envelope? Please update your question. Commented Dec 14, 2016 at 16:44
• I have edited you question. I assume you wanted to start the displayed expression with a binomial coefficient. If so, you can look at my coding to see how to do that. And to get $(1)^{1/3}$, you need to type (1)^{1/3}: only the first character after ^ goes into the superscript unless you enclose the entire superscript in braces. That said, I do not follow your logic for this expression. // Please re-edit your question if you meant something else. Commented Dec 14, 2016 at 17:02

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...

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.

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 .