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mathmath333
 one year ago
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
 one year ago
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

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1Three 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you are asking about the probability of proper letter for three persons or only one person ? means minimum probability or maximum ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1probability that at least one letter is in its proper envelope

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the probability to get a proper letter for each person is 3/9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0becz there are three letters and the total probability of these letters to deliver is 9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and the probability to get a proper letter for each person is 1/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01/3 because that is 3/9 in simplest form

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0plus its in simplest form

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Required 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
 one year ago
Best ResponseYou've already chosen the best response.1The total number of ways we could kee=6

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2It 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439671593426:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3trick question : whats the probability for having exactly two correct envelopes ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2At least one correct envelope = negation of (not even a single correct envelope)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2We can have not a single correct envelope in two ways.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So you got how there are in totality six ways to place the numbers in the envelopes?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1no i didnt got it yet

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3lets consider the group of 6 permutations

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439671878898:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3the identity element is : dw:1439671958498:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1ok i got it now 3!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Next consider the permutations that fix exactly one element : dw:1439672049158:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3in how many permutations do we have exactly one element fixed ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey mathmath333 can u fan me

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1u got 6 by 3P3 right ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Easy, we could fix x, or y, or z. so there are 3 permutations that fix exactly one element : dw:1439672237639:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what u mean mathmath333

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1the total number of ways we could keep them =6 is got by 3P3 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how many ways can you arrange "n" letters ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how many ways can you arrange "3' letters ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but in question it is different "how many ways can you arrange "n" letters in "m" different boxes ?"

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.13 letters are arranged in 3!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.33 letters and 3 people so you're right, it is 3P3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03p3 meaning 3 per person

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large ^{3}P_{3}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i 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."

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there 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
 one year ago
Best ResponseYou've already chosen the best response.3do you get this diagram that i took so much labor to draw ? :) dw:1439672841621:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and there we have to find the probability to proper letter so it will be 3/9=1/3

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1yes u described the six ways in diagram dw:1439672940726:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3nope, let me explain it again

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1439673019643:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3observe 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
 one year ago
Best ResponseYou've already chosen the best response.1ok so each in 4 types got at least one letter correct

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large \begin{matrix} \color{green}x&\color{green}y&\color{green}z\\ \color{green}x&\color{red}z&\color{red}y\\ \color{red}y&\color{red}x&\color{green}z\\ \color{red}y&\color{red}z&\color{red}x\\ \color{red}z&\color{red}x&\color{red}y\\ z&\color{green}y&x \end{matrix}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1439673430325:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3`ok so each in 4 types got at least one letter correct` looks you have figured it out!

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1u indicated persons and letters both with x,y,z that made confusing

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3get used to it, using different letters for persons and letters would be more confusing

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439673747997:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3sure, if that is easy for you :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3keeping track of 6 variables is something i totally avoid though..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3btw, there should be 6 permutations in total, we only see 4 in above diagram... see if you can sketch the remaining

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3above diagram covers permutations that : 1) fix exactly 3 elements 2) fix exactly 1 element what else are missing ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ganish 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 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now if there are 3 permutation for each then total permutation should be 3+3+3=9 isn't it ?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 was really good to show those 4 fov cases

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439674204791:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3total permutations of 3 letters is 3!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Awesome! what do you notice about those 2 last permutations ?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i meant that diagram

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1none of them got the correct letters to correct owners

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about about persons ? who are getting letters what will be their probability to get letter? ganesh

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3good, coming back to our problem, so how many permutations are in our favor in that list of 6 ?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.04/6 like path did

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes! 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 :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1439674657357:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3In abstract algebra that set with those 6 permutations is an important one. It forms a group under the operation, composition

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0let'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
 one year ago
Best ResponseYou've already chosen the best response.0is notgonna be permissible *
