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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.

Mathematics
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This is what I got but i dont know why its not right.. \[\tbinom{6}{3} *(0.39)^3+(0.61)^3 + \binom{4}{1} *(0.39)^1*(0.61)^3\]
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)
This is called Binomial Probability related to the Binomial Expansion..

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So your expression should be: \[\large (0.61)^3\times (0.39)^3 \times (0.80) \times (0.20)^3\]
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.
yeah, and i dont think your answer is right..
the binomial expansion: \[(a+b)^n = \binom{n}{r} *(a)^{n-r} * (b)^{n}\]
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..
nope..
\[\binom{6}{3} *(0.61)^3*(0.39)+\binom{4}{1} *(0.80)^1*(0.20)^3\] apparently not right
you obviously need to multiply the two probabilities, not add them - think about it.
still not right
close though
the answer is \(0.0060\)
and what do you seem to be getting?
\(0.0453\) man i hate probability; forget it
I am starting to hate it too -...-
lol its horrible; too confusing -_-

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