mathmath333
  • 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
Mathematics
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • 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
  • sohailiftikhar
you are asking about the probability of proper letter for three persons or only one person ? means minimum probability or maximum ?
mathmath333
  • mathmath333
probability that at least one letter is in its proper envelope

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More answers

sohailiftikhar
  • sohailiftikhar
the probability to get a proper letter for each person is 3/9
sohailiftikhar
  • sohailiftikhar
means 1/3
mathmath333
  • mathmath333
how
sohailiftikhar
  • sohailiftikhar
becz there are three letters and the total probability of these letters to deliver is 9
sohailiftikhar
  • sohailiftikhar
and the probability to get a proper letter for each person is 1/3
anonymous
  • anonymous
1/3 because that is 3/9 in simplest form
sohailiftikhar
  • sohailiftikhar
get it or not ?
anonymous
  • anonymous
i get it
sohailiftikhar
  • sohailiftikhar
good
anonymous
  • anonymous
plus its in simplest form
ParthKohli
  • 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
  • mathmath333
how u got 6
mathmath333
  • mathmath333
The total number of ways we could kee=6
ParthKohli
  • 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
  • ParthKohli
|dw:1439671593426:dw|
ganeshie8
  • ganeshie8
trick question : whats the probability for having exactly two correct envelopes ?
ParthKohli
  • ParthKohli
Zero.
ParthKohli
  • ParthKohli
At least one correct envelope = negation of (not even a single correct envelope)
sohailiftikhar
  • sohailiftikhar
yes
ParthKohli
  • ParthKohli
We can have not a single correct envelope in two ways.
mathmath333
  • mathmath333
how u got 4
ParthKohli
  • ParthKohli
So you got how there are in totality six ways to place the numbers in the envelopes?
mathmath333
  • mathmath333
no i didnt got it yet
ganeshie8
  • ganeshie8
lets consider the group of 6 permutations
ParthKohli
  • ParthKohli
|dw:1439671878898:dw|
ganeshie8
  • ganeshie8
the identity element is : |dw:1439671958498:dw|
mathmath333
  • mathmath333
ok i got it now 3!
ganeshie8
  • ganeshie8
Next consider the permutations that fix exactly one element : |dw:1439672049158:dw|
ganeshie8
  • ganeshie8
in how many permutations do we have exactly one element fixed ?
anonymous
  • anonymous
hey mathmath333 can u fan me
mathmath333
  • mathmath333
u got 6 by 3P3 right ?
ganeshie8
  • ganeshie8
Easy, we could fix x, or y, or z. so there are 3 permutations that fix exactly one element : |dw:1439672237639:dw|
anonymous
  • anonymous
what u mean mathmath333
mathmath333
  • mathmath333
the total number of ways we could keep them =6 is got by 3P3 ?
ganeshie8
  • ganeshie8
how many ways can you arrange "n" letters ?
mathmath333
  • mathmath333
n!
ganeshie8
  • ganeshie8
how many ways can you arrange "3' letters ?
mathmath333
  • mathmath333
but in question it is different "how many ways can you arrange "n" letters in "m" different boxes ?"
sohailiftikhar
  • sohailiftikhar
lol
mathmath333
  • mathmath333
3 letters are arranged in 3!
ganeshie8
  • ganeshie8
3 letters and 3 people so you're right, it is 3P3
mathmath333
  • mathmath333
ok
anonymous
  • anonymous
3p3 meaning 3 per person
mathmath333
  • mathmath333
\(\Large ^{3}P_{3}\)
mathmath333
  • 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
  • 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
  • ganeshie8
do you get this diagram that i took so much labor to draw ? :) |dw:1439672841621:dw|
sohailiftikhar
  • sohailiftikhar
and there we have to find the probability to proper letter so it will be 3/9=1/3
mathmath333
  • mathmath333
yes u described the six ways in diagram |dw:1439672940726:dw|
ganeshie8
  • ganeshie8
nope, let me explain it again
anonymous
  • anonymous
yes u did
sohailiftikhar
  • sohailiftikhar
ok
ganeshie8
  • ganeshie8
|dw:1439673019643:dw|
anonymous
  • anonymous
yes
ganeshie8
  • 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
  • mathmath333
ok so each in 4 types got at least one letter correct
amistre64
  • 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
  • ganeshie8
|dw:1439673430325:dw|
ganeshie8
  • ganeshie8
`ok so each in 4 types got at least one letter correct` looks you have figured it out!
mathmath333
  • mathmath333
u indicated persons and letters both with x,y,z that made confusing
ganeshie8
  • ganeshie8
get used to it, using different letters for persons and letters would be more confusing
mathmath333
  • mathmath333
|dw:1439673747997:dw|
ganeshie8
  • ganeshie8
sure, if that is easy for you :)
mathmath333
  • mathmath333
thnx :)
ganeshie8
  • ganeshie8
keeping track of 6 variables is something i totally avoid though..
ganeshie8
  • ganeshie8
btw, there should be 6 permutations in total, we only see 4 in above diagram... see if you can sketch the remaining
ganeshie8
  • ganeshie8
above diagram covers permutations that : 1) fix exactly 3 elements 2) fix exactly 1 element what else are missing ?
sohailiftikhar
  • 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
  • sohailiftikhar
now if there are 3 permutation for each then total permutation should be 3+3+3=9 isn't it ?
xapproachesinfinity
  • xapproachesinfinity
@ganeshie8 was really good to show those 4 fov cases
mathmath333
  • mathmath333
|dw:1439674204791:dw|
ganeshie8
  • ganeshie8
total permutations of 3 letters is 3!
sohailiftikhar
  • sohailiftikhar
yes he is :)
ganeshie8
  • ganeshie8
Awesome! what do you notice about those 2 last permutations ?
xapproachesinfinity
  • xapproachesinfinity
i meant that diagram
mathmath333
  • mathmath333
none of them got the correct letters to correct owners
sohailiftikhar
  • sohailiftikhar
what about about persons ? who are getting letters what will be their probability to get letter? ganesh
ganeshie8
  • ganeshie8
good, coming back to our problem, so how many permutations are in our favor in that list of 6 ?
mathmath333
  • mathmath333
4
xapproachesinfinity
  • xapproachesinfinity
4/6 like path did
ganeshie8
  • 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
  • sohailiftikhar
yes!
ganeshie8
  • ganeshie8
|dw:1439674657357:dw|
ganeshie8
  • ganeshie8
In abstract algebra that set with those 6 permutations is an important one. It forms a group under the operation, composition
xapproachesinfinity
  • 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
  • xapproachesinfinity
is notgonna be permissible *
ganeshie8
  • 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
  • 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
  • xapproachesinfinity
i do i agree with you no doubt :)

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