# Thread: Probability question

1. ## Probability question

A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E.

Find the probability that both of the letters to A and B are in incorrect envelopes.

The answer is given as 13/20 but I can't figure out why. Can anyone help?

2. Originally Posted by jjc
A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E. Find the probability that both of the letters to A and B are in incorrect envelopes. The answer is given as 13/20
The number of ways that A or B is in the correct envelope is give by:
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{{2(4!) - 3!}}{{5!}}$
Now, for neither to get the correct letter subtract from 1.

3. Hello, jjc!

A man writes 5 letters, one each to A, B, C, D and E.
Each letter is placed in a separate envelope and sealed.
He then addressed the envelopes, at random, one each to A, B, C, D and E.

Find the probability that both of the letters to A and B are in incorrect envelopes.

Answer: $\frac{13}{20}$
Here is a logical (but very primitive) approach . . .

The letters are: . $A,B,C,D,E$

Their envelopes are: . $a,b,c,d,e$

There are two cases to consider:
. . (1) Letter $A$ is in envelope $b.$
. . (2) Letter $A$ is not in envelope $b.$

(1) Letter $A$ is in envelope $b$: . $P(\text{A in b}) \:=\:\frac{1}{5}$

Then letter $B$ can go in any of the other four envelopes: $\{a,c,d,e\}$
. . $P(\text{B wrong}) \:=\:\frac{4}{4} \:=\:1$

Hence: . $P(\text{A in b }\wedge\text{ B wrong}) \:=\:\frac{1}{5}\cdot1 \:=\:{\color{blue}\frac{1}{5}}$

(2) Letter $A$ is in not in envelope $b$ ... it is in $\{c,d,e\}$
. . $P(\text{A not in b}) \:=\:\frac{3}{5}$

Then letter B must be in one of the three remaining envelopes.
. . $P(\text{B wrong}) \:=\:\frac{3}{4}$

Hence: . $P(\text{A not in b} \wedge\text{ B wrong}) \:=\:\frac{3}{5}\cdot\frac{3}{4} \:=\:{\color{blue}\frac{9}{20}}$

Therefore: . $P(\text{A wrong }\wedge\text{ B wrong}) \;=\;\frac{1}{5} + \frac{9}{20} \;=\;{\bf{\color{red}\frac{13}{20}}}$