my uncle wrote three different letters and addressed three envelopes. then he went outside for a walk. while he was out, his little daughter put a letter into each envelope and sealed it. what is the probability that none of the letters was in the correct envelope?

Guest Jan 16, 2015

#7**+13 **

Hi Rosala,

here all possibilities:

**I.**

Letter 1 put into envelope 1

Letter 2 put into envelope 2 or Letter 2 put into envelope 3

Letter 3 must be in envelope 3 Letter 3 must be in envelope 2

we have:

1. Letter 1 in envelope 1

Letter 2 in envelope 2

Letter 3 in envelope 3

2. Letter 1 in envelope 1

Letter 2 in envelope 3

Letter 3 in envelope 2

we can change places 3 times. Letter 1 can start in envelope 1 or in envelope 2 or in envelope 3

so we have 2 * 3 = 6

heureka
Jan 16, 2015

#1**+8 **

**my uncle wrote three different letters and addressed three envelopes. then he went outside for a walk. while he was out, his little daughter put a letter into each envelope and sealed it. what is the probability that none of the letters was in the correct envelope?**

$$\small{\text{

\textcolor[rgb]{0,1,0}{okay} \text{ \textcolor[rgb]{1,0,0}{false}

\begin{array}{|l|c|c|c|}

\hline

$n& letter 1 & letter 2 & letter 3 $ \\

\hline

$1& envelope\ \textcolor[rgb]{0,1,0}{1} & envelope\ \textcolor[rgb]{0,1,0}{2} & envelope\ \textcolor[rgb]{0,1,0}{3} $ \\

$2& envelope\ \textcolor[rgb]{0,1,0}{1} & envelope\ \textcolor[rgb]{1,0,0}{3} & envelope\ \textcolor[rgb]{1,0,0}{2} $ \\

$3& envelope\ \textcolor[rgb]{1,0,0}{2} & envelope\ \textcolor[rgb]{1,0,0}{1} & envelope\ \textcolor[rgb]{0,1,0}{3} $ \\

$\textcolor[rgb]{1,0,0}{4}& envelope\ \textcolor[rgb]{1,0,0}{2} & envelope\ \textcolor[rgb]{1,0,0}{3} & envelope\ \textcolor[rgb]{1,0,0}{1} $ \\

$\textcolor[rgb]{1,0,0}{5}& envelope\ \textcolor[rgb]{1,0,0}{3} & envelope\ \textcolor[rgb]{1,0,0}{1} & envelope\ \textcolor[rgb]{1,0,0}{2} $ \\

$6& envelope\ \textcolor[rgb]{1,0,0}{3} & envelope\ \textcolor[rgb]{0,1,0}{2} & envelope\ \textcolor[rgb]{1,0,0}{1} $ \\

\hline

\end{array}

}}$$

The probability that none of the letters was in the correct envelope is $$\frac{2}{6} = \frac{1}{3} = 33.\overline{3}\ \%$$

heureka
Jan 16, 2015

#2**+13 **

heureka....i dont understand you answer...i tried solving this but i couldnt.....can u explain your answer to me please!

rosala
Jan 16, 2015

#3**+13 **

Hi Rosala,

There are 6 different possibilities to put 3 letters into 3 envelopes. I have numbered serially them with from 1 to 6. Then I have looked which letters agree with which envelopes. There remain 2 possibilities in those all 3 letters in the envelopes are wrong franked. Then the probability is: 2/6

heureka
Jan 16, 2015

#4**+13 **

how come 6, it said ' what is the probability that none of the letters was in the correct envelope' , there are 3 letters and 3 envelopes so it should be 9.......im sooo confused Heureka!

rosala
Jan 16, 2015

#5**+13 **

Hi Rosala,

for example if Letter 1 is in envelope 1 then you can't put Letter 2 or Letter 3 into envelope 1.

heureka
Jan 16, 2015

#6**+13 **

hi heaureka,...so now i understood why not 9 ...but would you mind telling me why6?im just so confused thats why!thanks!

rosala
Jan 16, 2015

#7**+13 **

Best Answer

Hi Rosala,

here all possibilities:

**I.**

Letter 1 put into envelope 1

Letter 2 put into envelope 2 or Letter 2 put into envelope 3

Letter 3 must be in envelope 3 Letter 3 must be in envelope 2

we have:

1. Letter 1 in envelope 1

Letter 2 in envelope 2

Letter 3 in envelope 3

2. Letter 1 in envelope 1

Letter 2 in envelope 3

Letter 3 in envelope 2

we can change places 3 times. Letter 1 can start in envelope 1 or in envelope 2 or in envelope 3

so we have 2 * 3 = 6

heureka
Jan 16, 2015