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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

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  1. mathmath333
    • one year ago
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    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. sohailiftikhar
    • one year ago
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    you are asking about the probability of proper letter for three persons or only one person ? means minimum probability or maximum ?

  3. mathmath333
    • one year ago
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    probability that at least one letter is in its proper envelope

  4. sohailiftikhar
    • one year ago
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    the probability to get a proper letter for each person is 3/9

  5. sohailiftikhar
    • one year ago
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    means 1/3

  6. mathmath333
    • one year ago
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    how

  7. sohailiftikhar
    • one year ago
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    becz there are three letters and the total probability of these letters to deliver is 9

  8. sohailiftikhar
    • one year ago
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    and the probability to get a proper letter for each person is 1/3

  9. anonymous
    • one year ago
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    1/3 because that is 3/9 in simplest form

  10. sohailiftikhar
    • one year ago
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    get it or not ?

  11. anonymous
    • one year ago
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    i get it

  12. sohailiftikhar
    • one year ago
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    good

  13. anonymous
    • one year ago
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    plus its in simplest form

  14. ParthKohli
    • one year ago
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    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
    • one year ago
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    how u got 6

  16. mathmath333
    • one year ago
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    The total number of ways we could kee=6

  17. ParthKohli
    • one year ago
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    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
    • one year ago
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    |dw:1439671593426:dw|

  19. ganeshie8
    • one year ago
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    trick question : whats the probability for having exactly two correct envelopes ?

  20. ParthKohli
    • one year ago
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    Zero.

  21. ParthKohli
    • one year ago
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    At least one correct envelope = negation of (not even a single correct envelope)

  22. sohailiftikhar
    • one year ago
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    yes

  23. ParthKohli
    • one year ago
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    We can have not a single correct envelope in two ways.

  24. mathmath333
    • one year ago
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    how u got 4

  25. ParthKohli
    • one year ago
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    So you got how there are in totality six ways to place the numbers in the envelopes?

  26. mathmath333
    • one year ago
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    no i didnt got it yet

  27. ganeshie8
    • one year ago
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    lets consider the group of 6 permutations

  28. ParthKohli
    • one year ago
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    |dw:1439671878898:dw|

  29. ganeshie8
    • one year ago
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    the identity element is : |dw:1439671958498:dw|

  30. mathmath333
    • one year ago
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    ok i got it now 3!

  31. ganeshie8
    • one year ago
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    Next consider the permutations that fix exactly one element : |dw:1439672049158:dw|

  32. ganeshie8
    • one year ago
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    in how many permutations do we have exactly one element fixed ?

  33. anonymous
    • one year ago
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    hey mathmath333 can u fan me

  34. mathmath333
    • one year ago
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    u got 6 by 3P3 right ?

  35. ganeshie8
    • one year ago
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    Easy, we could fix x, or y, or z. so there are 3 permutations that fix exactly one element : |dw:1439672237639:dw|

  36. anonymous
    • one year ago
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    what u mean mathmath333

  37. mathmath333
    • one year ago
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    the total number of ways we could keep them =6 is got by 3P3 ?

  38. ganeshie8
    • one year ago
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    how many ways can you arrange "n" letters ?

  39. mathmath333
    • one year ago
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    n!

  40. ganeshie8
    • one year ago
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    how many ways can you arrange "3' letters ?

  41. mathmath333
    • one year ago
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    but in question it is different "how many ways can you arrange "n" letters in "m" different boxes ?"

  42. sohailiftikhar
    • one year ago
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    lol

  43. mathmath333
    • one year ago
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    3 letters are arranged in 3!

  44. ganeshie8
    • one year ago
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    3 letters and 3 people so you're right, it is 3P3

  45. mathmath333
    • one year ago
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    ok

  46. anonymous
    • one year ago
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    3p3 meaning 3 per person

  47. mathmath333
    • one year ago
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    \(\Large ^{3}P_{3}\)

  48. mathmath333
    • one year ago
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    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. sohailiftikhar
    • one year ago
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    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
    • one year ago
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    do you get this diagram that i took so much labor to draw ? :) |dw:1439672841621:dw|

  51. sohailiftikhar
    • one year ago
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    and there we have to find the probability to proper letter so it will be 3/9=1/3

  52. mathmath333
    • one year ago
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    yes u described the six ways in diagram |dw:1439672940726:dw|

  53. ganeshie8
    • one year ago
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    nope, let me explain it again

  54. anonymous
    • one year ago
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    yes u did

  55. sohailiftikhar
    • one year ago
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    ok

  56. ganeshie8
    • one year ago
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    |dw:1439673019643:dw|

  57. anonymous
    • one year ago
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    yes

  58. ganeshie8
    • one year ago
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    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
    • one year ago
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    ok so each in 4 types got at least one letter correct

  60. amistre64
    • one year ago
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    \[\Large \begin{matrix} \color{green}x&\color{green}y&\color{green}z\\ \color{green}x&\color{red}z&\color{red}y\\ \color{red}y&\color{red}x&\color{green}z\\ \color{red}y&\color{red}z&\color{red}x\\ \color{red}z&\color{red}x&\color{red}y\\ z&\color{green}y&x \end{matrix}\]

  61. ganeshie8
    • one year ago
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    |dw:1439673430325:dw|

  62. ganeshie8
    • one year ago
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    `ok so each in 4 types got at least one letter correct` looks you have figured it out!

  63. mathmath333
    • one year ago
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    u indicated persons and letters both with x,y,z that made confusing

  64. ganeshie8
    • one year ago
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    get used to it, using different letters for persons and letters would be more confusing

  65. mathmath333
    • one year ago
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    |dw:1439673747997:dw|

  66. ganeshie8
    • one year ago
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    sure, if that is easy for you :)

  67. mathmath333
    • one year ago
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    thnx :)

  68. ganeshie8
    • one year ago
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    keeping track of 6 variables is something i totally avoid though..

  69. ganeshie8
    • one year ago
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    btw, there should be 6 permutations in total, we only see 4 in above diagram... see if you can sketch the remaining

  70. ganeshie8
    • one year ago
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    above diagram covers permutations that : 1) fix exactly 3 elements 2) fix exactly 1 element what else are missing ?

  71. sohailiftikhar
    • one year ago
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    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. sohailiftikhar
    • one year ago
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    now if there are 3 permutation for each then total permutation should be 3+3+3=9 isn't it ?

  73. xapproachesinfinity
    • one year ago
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    @ganeshie8 was really good to show those 4 fov cases

  74. mathmath333
    • one year ago
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    |dw:1439674204791:dw|

  75. ganeshie8
    • one year ago
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    total permutations of 3 letters is 3!

  76. sohailiftikhar
    • one year ago
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    yes he is :)

  77. ganeshie8
    • one year ago
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    Awesome! what do you notice about those 2 last permutations ?

  78. xapproachesinfinity
    • one year ago
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    i meant that diagram

  79. mathmath333
    • one year ago
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    none of them got the correct letters to correct owners

  80. sohailiftikhar
    • one year ago
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    what about about persons ? who are getting letters what will be their probability to get letter? ganesh

  81. ganeshie8
    • one year ago
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    good, coming back to our problem, so how many permutations are in our favor in that list of 6 ?

  82. mathmath333
    • one year ago
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    4

  83. xapproachesinfinity
    • one year ago
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    4/6 like path did

  84. ganeshie8
    • one year ago
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    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. sohailiftikhar
    • one year ago
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    yes!

  86. ganeshie8
    • one year ago
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    |dw:1439674657357:dw|

  87. ganeshie8
    • one year ago
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    In abstract algebra that set with those 6 permutations is an important one. It forms a group under the operation, composition

  88. xapproachesinfinity
    • one year ago
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    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
    • one year ago
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    is notgonna be permissible *

  90. ganeshie8
    • one year ago
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    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..