## 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

1. 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.

2. anonymous

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

3. mathmath333

probability that at least one letter is in its proper envelope

4. anonymous

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

5. anonymous

means 1/3

6. mathmath333

how

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

get it or not ?

11. anonymous

i get it

12. anonymous

good

13. anonymous

plus its in simplest form

14. 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.

15. mathmath333

how u got 6

16. mathmath333

The total number of ways we could kee=6

17. 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.

18. ParthKohli

|dw:1439671593426:dw|

19. ganeshie8

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

20. ParthKohli

Zero.

21. ParthKohli

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

22. anonymous

yes

23. ParthKohli

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

24. mathmath333

how u got 4

25. ParthKohli

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

26. mathmath333

no i didnt got it yet

27. ganeshie8

lets consider the group of 6 permutations

28. ParthKohli

|dw:1439671878898:dw|

29. ganeshie8

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

30. mathmath333

ok i got it now 3!

31. ganeshie8

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

32. ganeshie8

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

33. anonymous

hey mathmath333 can u fan me

34. mathmath333

u got 6 by 3P3 right ?

35. ganeshie8

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

36. anonymous

what u mean mathmath333

37. mathmath333

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

38. ganeshie8

how many ways can you arrange "n" letters ?

39. mathmath333

n!

40. ganeshie8

how many ways can you arrange "3' letters ?

41. mathmath333

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

42. anonymous

lol

43. mathmath333

3 letters are arranged in 3!

44. ganeshie8

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

45. mathmath333

ok

46. anonymous

3p3 meaning 3 per person

47. mathmath333

$$\Large ^{3}P_{3}$$

48. 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."

49. anonymous

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

50. ganeshie8

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

51. anonymous

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

52. mathmath333

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

53. ganeshie8

nope, let me explain it again

54. anonymous

yes u did

55. anonymous

ok

56. ganeshie8

|dw:1439673019643:dw|

57. anonymous

yes

58. 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

59. mathmath333

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

60. amistre64

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

61. ganeshie8

|dw:1439673430325:dw|

62. ganeshie8

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

63. mathmath333

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

64. ganeshie8

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

65. mathmath333

|dw:1439673747997:dw|

66. ganeshie8

sure, if that is easy for you :)

67. mathmath333

thnx :)

68. ganeshie8

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

69. ganeshie8

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

70. ganeshie8

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

71. anonymous

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 ?

72. anonymous

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

73. xapproachesinfinity

@ganeshie8 was really good to show those 4 fov cases

74. mathmath333

|dw:1439674204791:dw|

75. ganeshie8

total permutations of 3 letters is 3!

76. anonymous

yes he is :)

77. ganeshie8

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

78. xapproachesinfinity

i meant that diagram

79. mathmath333

none of them got the correct letters to correct owners

80. anonymous

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

81. ganeshie8

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

82. mathmath333

4

83. xapproachesinfinity

4/6 like path did

84. 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 :)

85. anonymous

yes!

86. ganeshie8

|dw:1439674657357:dw|

87. ganeshie8

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

88. 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

89. xapproachesinfinity

is notgonna be permissible *