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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

Best Answer 

 #7
avatar+19995 
+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
avatar+19995 
+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
avatar+11847 
+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
avatar+19995 
+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
avatar+11847 
+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
avatar+19995 
+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
avatar+11847 
+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
avatar+19995 
+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
 #8
avatar+93299 
+8

Thanks Heureka  

And thanks for the good questions Rosala.    

Melody  Jan 16, 2015
 #9
avatar+11847 
+8

Thanks Melody and Heureka!

 

Thank you Heureka for all the hard work for me!i really appreciate it!thanks!

 

rosala  Jan 16, 2015

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