# probability - urgent help please

• Oct 16th 2007, 04:09 AM
simone
I'm having lots of problems with probability questions.
Could you help me solve this? Thank you.

4 different letters are to be sent to 4 different addresses. For each letter there is an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what it the probability that ony 1 letter will be put into the envelope with its correct address?

I thought each letter has a 1/4 chance of being put in the right envelope, and a 3/4 chance of being put in the wrong envelope, so
P=(1/4)(3/4)(3/4)(3/4)

what am I doing wrong?

thank youhttp://www.mathisfunforum.com/img/smilies/wave.gif
• Oct 16th 2007, 05:45 AM
Plato
Say that the letters are A,B,C, & D. Write out all 24 permutations of ABCD. You will see that there are only two cases in which A alone is in its correct place. You will also find that there only cases two having B alone in its correct place. The same is true for C&D.
ABCD
ABDC
ACBD
ACDB
BACD
BCDA
BDAC
BDCA
CABD

CBDA
CDAB
CDBA
DABC
DACB
DBAC

DBCA
DCAB
DCBA
• Oct 16th 2007, 06:02 AM
simone
Thank you Plato, I got the concept. Could you please tell me how to put all this in a mathematical formula without having to list all the permutations and having to sort through them?
Thank you so much.
(Doh)
• Oct 16th 2007, 08:20 AM
Soroban
Hello, Simone!

Quote:

Four different letters are to be sent to 4 different addresses.
For each letter there is an envelope with its correct address.
If the 4 letters are to be put into the 4 envelopes at random,
what is the probability that ony 1 letter will be put into the envelope with its correct address?

There are: .$\displaystyle 4! = {\color{blue}24}$ ways to place the letters.

Suppose the letters are: .$\displaystyle \{A,B,C,D\}$
. . and their corresponding envelopes are: .$\displaystyle \{a,b,c,d\}$

One letter is in its correct envelope. .There are four choices for this.

The other three must not be in their corrsponding envelopes.
. . In how many ways can this happen?

Suppose $\displaystyle A$ is in envelope $\displaystyle a.$ .Then $\displaystyle B,C,D$ must be the wrong envelopes.

There are only two ways: . $\displaystyle \begin{array}{ccc}b&c&d\\ \hline C&D&B\\D&B&C\end{array}$

Hence, there are: .$\displaystyle 4 \times 2 \:=\:{\color{blue}8}$ ways to have one letter correct.

Therefore: .$\displaystyle P(\text{one correct}) \;=\;\frac{8}{24} \;=\;\boxed{\frac{1}{3}}$

• Oct 16th 2007, 08:20 AM
Plato
Quote:

Originally Posted by simone
Could you please tell me how to put all this in a mathematical formula without having to list all the permutations and having to sort through them.

The answer depends on how much counting theory one knows?
This is purely a problem in counting.