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mathmath333

  • one year ago

Probability Question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& \normalsize \text{In a relay race there are five teams A, B, C, D and E.}\hspace{.33em}\\~\\ & \normalsize \text{(a) What is the probability that A, B and C}\hspace{.33em}\\~\\ & \normalsize \text{are first three to finish (in any order)}\hspace{.33em}\\~\\ & \normalsize \text{(Assume that all finishing orders are equally likely)}\hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    whats your first step?

  3. ganeshie8
    • one year ago
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    lets do this problem in two ways : method 1 we can choose any \(3\) people from \(5\) people in \(\large ^5C_3\) ways, but only the selection \(\{A,B,C\}\) is our favorite, so the probability is \(\large \dfrac{1}{^5C_3}\)

  4. ganeshie8
    • one year ago
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    method2 : consider a string of length \(3\) : |dw:1439647639858:dw| how many total strings of length 3 can you make by using the letters \(\{A,B,C,D,E\}\) ?

  5. mathmath333
    • one year ago
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    \(\large 3^{3}\) ?

  6. ganeshie8
    • one year ago
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    Nope, how many choices are there for the first letter ? |dw:1439647853502:dw|

  7. mathmath333
    • one year ago
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    5

  8. mathmath333
    • one year ago
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    5*4*3=60

  9. ganeshie8
    • one year ago
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    Yes, thats the total number of strings. How many of those strings have A,B,C in any order ?

  10. mathmath333
    • one year ago
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    6

  11. ganeshie8
    • one year ago
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    In other words, how many 3 bit strings can you make using just the letters A,B,C

  12. ganeshie8
    • one year ago
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    6 is right

  13. ganeshie8
    • one year ago
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    total number of strings = 60 number of strings in favor = 6 divide to get the probability

  14. mathmath333
    • one year ago
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    thnx!

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