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Mimi_x3 Group Title

Probability: A man is restoring ten old cars, six of them manufactured in \(1955\) and four of them manufactured in \(1962\). When he tries to start them, on average the \(1955\) models will start \(65\) percent of the time and the 1962 models will start \(80\) percent of time. Find the probability that any time: a) exactly three of the \(1955\) models and one of the \(1962\) model will start.

  • 2 years ago
  • 2 years ago

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  1. Mimi_x3 Group Title
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    This is what I got but i dont know why its not right.. \[\tbinom{6}{3} *(0.39)^3+(0.61)^3 + \binom{4}{1} *(0.39)^1*(0.61)^3\]

    • 2 years ago
  2. apoorvk Group Title
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    Okay yes, so why did you multiply those factorials - He does not have to 'choose' cars - the cars that start up could be any three outta the 6! And also, you multiply together the probabilities for both kinds of cars - since in the end you want 4 cars to start and 6 not to (with their own respective probabilities governing their 'fates'). (and I believe you wanted to 'multiply' the cube of 0.61 over there, and meant '0.20' and '0.80' for the 1962 models)

    • 2 years ago
  3. Mimi_x3 Group Title
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    This is called Binomial Probability related to the Binomial Expansion..

    • 2 years ago
  4. apoorvk Group Title
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    So your expression should be: \[\large (0.61)^3\times (0.39)^3 \times (0.80) \times (0.20)^3\]

    • 2 years ago
  5. apoorvk Group Title
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    Yeah I was guessing this was something like Binomial - but I don't know what use multiply the 'combinations' would be in this particular case.

    • 2 years ago
  6. Mimi_x3 Group Title
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    yeah, and i dont think your answer is right..

    • 2 years ago
  7. Mimi_x3 Group Title
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    the binomial expansion: \[(a+b)^n = \binom{n}{r} *(a)^{n-r} * (b)^{n}\]

    • 2 years ago
  8. apoorvk Group Title
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    Anyways, just correct the 'multiplication' sign in your equation and the '0.80' values, and see if that gets you the answer... I am still wondering..

    • 2 years ago
  9. Mimi_x3 Group Title
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    nope..

    • 2 years ago
  10. Mimi_x3 Group Title
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    \[\binom{6}{3} *(0.61)^3*(0.39)+\binom{4}{1} *(0.80)^1*(0.20)^3\] apparently not right

    • 2 years ago
  11. apoorvk Group Title
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    you obviously need to multiply the two probabilities, not add them - think about it.

    • 2 years ago
  12. Mimi_x3 Group Title
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    still not right

    • 2 years ago
  13. Mimi_x3 Group Title
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    close though

    • 2 years ago
  14. Mimi_x3 Group Title
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    the answer is \(0.0060\)

    • 2 years ago
  15. apoorvk Group Title
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    and what do you seem to be getting?

    • 2 years ago
  16. Mimi_x3 Group Title
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    \(0.0453\) man i hate probability; forget it

    • 2 years ago
  17. apoorvk Group Title
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    I am starting to hate it too -...-

    • 2 years ago
  18. Mimi_x3 Group Title
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    lol its horrible; too confusing -_-

    • 2 years ago
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