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mathmath333

  • one year ago

Probablity question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Three different numbers are selected from the set}\ X=\{1,2,3,4...10\}\hspace{.33em}\\~\\ & \normalsize \text{What is the probablity that the product of the two numbers is equal to }\hspace{.33em}\\~\\ & \normalsize \text{the third }\hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & a.)\ \dfrac{3}{10}\hspace{.33em}\\~\\ & b.)\ \dfrac{1}{40}\hspace{.33em}\\~\\ & c.)\ \dfrac{1}{20}\hspace{.33em}\\~\\ & d.)\ \dfrac{4}{5}\hspace{.33em}\\~\\ \end{align}}\)

  3. ganeshie8
    • one year ago
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    say the \(3\) numbers selected are \(\{a,b,c\}\)

  4. ganeshie8
    • one year ago
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    also suppose that \(a\lt b\lt c\)

  5. mathmath333
    • one year ago
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    ok

  6. ganeshie8
    • one year ago
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    then you want \(a*b\le 10\)

  7. ganeshie8
    • one year ago
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    notice that if \(a*b\le 10\), then \(a \le \sqrt{~10~}\)

  8. ganeshie8
    • one year ago
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    so \(a\) can only be either \(2\) or \(3\)

  9. ganeshie8
    • one year ago
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    when \(a=2\), check what all \(b\) values will work

  10. mathmath333
    • one year ago
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    how does the condition \(a=\sqrt{10}\) came.

  11. ganeshie8
    • one year ago
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    good question, that is called seive of eratosthenes https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

  12. mathmath333
    • one year ago
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    ok

  13. ganeshie8
    • one year ago
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    we have \(ab\le 10\) and you know that \(a\lt b\) whats the maximum value that \(a\) can take ?

  14. mathmath333
    • one year ago
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    2

  15. ganeshie8
    • one year ago
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    why not 3 ?

  16. mathmath333
    • one year ago
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    cuz \(3\times 4=12\)

  17. ganeshie8
    • one year ago
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    Perfect! so the only value that \(a\) can take is \(2\)

  18. mathmath333
    • one year ago
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    why not \(1\) ?

  19. ganeshie8
    • one year ago
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    you tell me why

  20. mathmath333
    • one year ago
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    \(1*2\le 10\)

  21. ganeshie8
    • one year ago
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    remember you want to pick all 3 "different" numbers

  22. ganeshie8
    • one year ago
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    what happens if any one number is \(1\) ?

  23. mathmath333
    • one year ago
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    it satisfies the condition example \(1,2,3\)

  24. ganeshie8
    • one year ago
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    it doesn't, 1*2 is not 3

  25. ganeshie8
    • one year ago
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    the smaller two numbers must multiply to the third number

  26. mathmath333
    • one year ago
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    oh i see, sry

  27. mathmath333
    • one year ago
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    yea , only \(2\) for \(a\)

  28. ganeshie8
    • one year ago
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    yes let \(a=2\) and find all \(b\) such that \(a*b\le 10\)

  29. mathmath333
    • one year ago
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    \(b=3,4,5\)

  30. ganeshie8
    • one year ago
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    Yes, so the numbers in favor are : (2, 3, 6) (2, 4, 8) (2, 5, 10)

  31. ganeshie8
    • one year ago
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    thats 3 in favor

  32. ganeshie8
    • one year ago
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    save that

  33. ganeshie8
    • one year ago
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    next, find how many total ways are there to choose 3 different numbers from the given 10 numbers

  34. mathmath333
    • one year ago
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    Is their any condition to choose 3 different numbers from the given 10 numbers.

  35. ganeshie8
    • one year ago
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    thats the sample space so no conditions, just find the total number of ways of choosing 3 numbers from 10

  36. mathmath333
    • one year ago
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    \(\dfrac{10!}{7!}\)

  37. ganeshie8
    • one year ago
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    nope

  38. ganeshie8
    • one year ago
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    its just \(\large \dbinom{10}{3}\)

  39. ganeshie8
    • one year ago
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    notice that here order doesn't matter... so it is a combination

  40. mathmath333
    • one year ago
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    why not \(^{10}P_{3}\) ??

  41. mathmath333
    • one year ago
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    ok

  42. ganeshie8
    • one year ago
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    favor : \(\large 3\) total : \(\large \dbinom{10}{3}\)

  43. ganeshie8
    • one year ago
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    take the ratio for the probability

  44. mathmath333
    • one year ago
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    so answer=\(\dfrac{1}{40}\)

  45. ganeshie8
    • one year ago
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    Yes, did u get why this is a combination and not a permutation problem ?

  46. mathmath333
    • one year ago
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    cuz we choosing 3 numbers one by one and not forming 3 numbers like \(abc_{10}\)

  47. ganeshie8
    • one year ago
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    yes, we never bothered about order while doing this problem

  48. ganeshie8
    • one year ago
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    we could also do it using permutations but it is painful here

  49. ganeshie8
    • one year ago
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    if we use permutations, the count of favor also changes

  50. ganeshie8
    • one year ago
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    (2, 3, 6) (2, 4, 8) (2, 5, 10)

  51. ganeshie8
    • one year ago
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    since (2, 3, 6) works, all the 6 permutations of it also work : (2, 6, 3) (3, 2, 6) (3, 6, 2) (6, 2, 3) (6, 3, 2)

  52. ganeshie8
    • one year ago
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    after all that mess, you will get the same answer

  53. ganeshie8
    • one year ago
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    In these probability problems, it doesn't matter whether you use permutations or combinations.. the final answer wont change if you do it correctly

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