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Mimi_x3

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.

  • one year ago
  • one year ago

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

    • one year ago
  2. apoorvk
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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)

    • one year ago
  3. Mimi_x3
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    This is called Binomial Probability related to the Binomial Expansion..

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

    • one year ago
  5. apoorvk
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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.

    • one year ago
  6. Mimi_x3
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    yeah, and i dont think your answer is right..

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

    • one year ago
  8. apoorvk
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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..

    • one year ago
  9. Mimi_x3
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    nope..

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

    • one year ago
  11. apoorvk
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    you obviously need to multiply the two probabilities, not add them - think about it.

    • one year ago
  12. Mimi_x3
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    still not right

    • one year ago
  13. Mimi_x3
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    close though

    • one year ago
  14. Mimi_x3
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    the answer is \(0.0060\)

    • one year ago
  15. apoorvk
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    and what do you seem to be getting?

    • one year ago
  16. Mimi_x3
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    \(0.0453\) man i hate probability; forget it

    • one year ago
  17. apoorvk
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    I am starting to hate it too -...-

    • one year ago
  18. Mimi_x3
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    lol its horrible; too confusing -_-

    • one year ago
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