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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.
 one year ago
 one year ago
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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Mimi_x3Best ResponseYou've already chosen the best response.2
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

apoorvkBest ResponseYou've already chosen the best response.1
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

Mimi_x3Best ResponseYou've already chosen the best response.2
This is called Binomial Probability related to the Binomial Expansion..
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
So your expression should be: \[\large (0.61)^3\times (0.39)^3 \times (0.80) \times (0.20)^3\]
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
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

Mimi_x3Best ResponseYou've already chosen the best response.2
yeah, and i dont think your answer is right..
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
the binomial expansion: \[(a+b)^n = \binom{n}{r} *(a)^{nr} * (b)^{n}\]
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
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

Mimi_x3Best ResponseYou've already chosen the best response.2
\[\binom{6}{3} *(0.61)^3*(0.39)+\binom{4}{1} *(0.80)^1*(0.20)^3\] apparently not right
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
you obviously need to multiply the two probabilities, not add them  think about it.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
the answer is \(0.0060\)
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
and what do you seem to be getting?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
\(0.0453\) man i hate probability; forget it
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
I am starting to hate it too ...
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
lol its horrible; too confusing _
 one year ago
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