Three letters are dictated to three persons and an envelope is addressed to each
of them, the letters are inserted into the envelopes at random so that each envelope
contains exactly one letter. Find the probability that at least one letter is in its
proper envelope

- mathmath333

- katieb

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- mathmath333

Three letters are dictated to three persons and an envelope is addressed to each
of them, the letters are inserted into the envelopes at random so that each envelope
contains exactly one letter. Find the probability that at least one letter is in its
proper envelope.

- sohailiftikhar

you are asking about the probability of proper letter for three persons or only one person ? means minimum probability or maximum ?

- mathmath333

probability that at least one letter is in its
proper envelope

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## More answers

- sohailiftikhar

the probability to get a proper letter for each person is 3/9

- sohailiftikhar

means 1/3

- mathmath333

how

- sohailiftikhar

becz there are three letters and the total probability of these letters to deliver is 9

- sohailiftikhar

and the probability to get a proper letter for each person is 1/3

- anonymous

1/3 because that is 3/9 in simplest form

- sohailiftikhar

get it or not ?

- anonymous

i get it

- sohailiftikhar

good

- anonymous

plus its in simplest form

- ParthKohli

Required probability is given by the number of ways the letters could be placed such that at least one letter is in its correct place divided by the number of ways we could place the letters (favourable ways/total ways).
The total number of ways we could keep them is \(3 \times 2 \times 1 = 6\).
One out of those ways is when all three are in the correct place.
Three other ways when exactly one is in its correct place.
Therefore, the number of favourable cases = 4.
Now we have P = 4/6 = 2/3.

- mathmath333

how u got 6

- mathmath333

The total number of ways we could kee=6

- ParthKohli

It makes sense that the probability should be 2/3 rather than 1/3. It's a lot more probable to keep at least one number in the right envelope.

- ParthKohli

|dw:1439671593426:dw|

- ganeshie8

trick question : whats the probability for having exactly two correct envelopes ?

- ParthKohli

Zero.

- ParthKohli

At least one correct envelope = negation of (not even a single correct envelope)

- sohailiftikhar

yes

- ParthKohli

We can have not a single correct envelope in two ways.

- mathmath333

how u got 4

- ParthKohli

So you got how there are in totality six ways to place the numbers in the envelopes?

- mathmath333

no i didnt got it yet

- ganeshie8

lets consider the group of 6 permutations

- ParthKohli

|dw:1439671878898:dw|

- ganeshie8

the identity element is :
|dw:1439671958498:dw|

- mathmath333

ok i got it now 3!

- ganeshie8

Next consider the permutations that fix exactly one element :
|dw:1439672049158:dw|

- ganeshie8

in how many permutations do we have exactly one element fixed ?

- anonymous

hey mathmath333 can u fan me

- mathmath333

u got 6 by 3P3 right ?

- ganeshie8

Easy, we could fix x, or y, or z. so there are 3 permutations that fix exactly one element :
|dw:1439672237639:dw|

- anonymous

what u mean mathmath333

- mathmath333

the total number of ways we could keep them =6 is got by 3P3 ?

- ganeshie8

how many ways can you arrange "n" letters ?

- mathmath333

n!

- ganeshie8

how many ways can you arrange "3' letters ?

- mathmath333

but in question it is different "how many ways can you arrange "n" letters in "m" different boxes ?"

- sohailiftikhar

lol

- mathmath333

3 letters are arranged in 3!

- ganeshie8

3 letters and 3 people
so you're right, it is 3P3

- mathmath333

ok

- anonymous

3p3 meaning 3 per person

- mathmath333

\(\Large ^{3}P_{3}\)

- mathmath333

i didn;t underdtand how parthkohli got 4 for numerator
"One out of those ways is when all three are in the correct place.
Three other ways when exactly one is in its correct place.
Therefore, the number of favourable cases = 4."

- sohailiftikhar

there is a total probability to give letter to three persons is 9
let the letters are x,y,z
now for person =p1
xyz
yxz
zyx
and for p2
xyz
yxz
zyx
same for p3

- ganeshie8

do you get this diagram that i took so much labor to draw ? :)
|dw:1439672841621:dw|

- sohailiftikhar

and there we have to find the probability to proper letter so it will be 3/9=1/3

- mathmath333

yes u described the six ways in diagram |dw:1439672940726:dw|

- ganeshie8

nope, let me explain it again

- anonymous

yes u did

- sohailiftikhar

ok

- ganeshie8

|dw:1439673019643:dw|

- anonymous

yes

- ganeshie8

observe the notation,
the left hand column indicates the letters,
the right hand column indicates the persons
the first permutation says
the letter x was sent to the person x
the letter y was sent to the person y
the letter z was sent to the person z
so all letters are sent to correct persons in this permutation

- mathmath333

ok so each in 4 types got at least one letter correct

- amistre64

\[\Large \begin{matrix}
\color{green}x&\color{green}y&\color{green}z\\
\color{green}x&\color{red}z&\color{red}y\\
\color{red}y&\color{red}x&\color{green}z\\
\color{red}y&\color{red}z&\color{red}x\\
\color{red}z&\color{red}x&\color{red}y\\
z&\color{green}y&x
\end{matrix}\]

- ganeshie8

|dw:1439673430325:dw|

- ganeshie8

`ok so each in 4 types got at least one letter correct`
looks you have figured it out!

- mathmath333

u indicated persons and letters both with x,y,z that made confusing

- ganeshie8

get used to it,
using different letters for persons and letters would be more confusing

- mathmath333

|dw:1439673747997:dw|

- ganeshie8

sure, if that is easy for you :)

- mathmath333

thnx :)

- ganeshie8

keeping track of 6 variables is something i totally avoid though..

- ganeshie8

btw, there should be 6 permutations in total, we only see 4 in above diagram... see if you can sketch the remaining

- ganeshie8

above diagram covers permutations that :
1) fix exactly 3 elements
2) fix exactly 1 element
what else are missing ?

- sohailiftikhar

ganish there is a 3 for each person look at your sketch if you are saying x as a person so there is 3 permutation for him like xx xy xz ...right ?

- sohailiftikhar

now if there are 3 permutation for each then total permutation should be 3+3+3=9 isn't it ?

- xapproachesinfinity

@ganeshie8 was really good to show those 4 fov cases

- mathmath333

|dw:1439674204791:dw|

- ganeshie8

total permutations of 3 letters is 3!

- sohailiftikhar

yes he is :)

- ganeshie8

Awesome! what do you notice about those 2 last permutations ?

- xapproachesinfinity

i meant that diagram

- mathmath333

none of them got the correct letters to correct owners

- sohailiftikhar

what about about persons ? who are getting letters what will be their probability to get letter? ganesh

- ganeshie8

good, coming back to our problem, so how many permutations are in our favor in that list of 6 ?

- mathmath333

4

- xapproachesinfinity

4/6 like path did

- ganeshie8

Yes! I'm trying to focus more on making u do it my way by listing out all permutations haha! I do see you got hang of it :)

- sohailiftikhar

yes!

- ganeshie8

|dw:1439674657357:dw|

- ganeshie8

In abstract algebra that set with those 6 permutations is an important one. It forms a group under the operation, composition

- xapproachesinfinity

let's say you have 100 people 100 letter and 100 envlopes
how would we find a Prob that at least 10 of them got the correct letters
that way of listing all the possibilities one by one is gonna be a permissible way

- xapproachesinfinity

is notgonna be permissible *

- ganeshie8

you need to know how to list them so that you get a feel of whats going on
simply using the formula directly is not a good thing..

- ganeshie8

for sure, that listing is a poor method for getting the answer...
it helps in understanding permutations so that the formulas make sense later

- xapproachesinfinity

i do i agree with you no doubt :)

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