Mimi_x3
  • 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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mimi_x3
  • Mimi_x3
This is what I got but i dont know why its not right.. \[\tbinom{6}{3} *(0.39)^3+(0.61)^3 + \binom{4}{1} *(0.39)^1*(0.61)^3\]
apoorvk
  • apoorvk
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)
Mimi_x3
  • Mimi_x3
This is called Binomial Probability related to the Binomial Expansion..

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

apoorvk
  • apoorvk
So your expression should be: \[\large (0.61)^3\times (0.39)^3 \times (0.80) \times (0.20)^3\]
apoorvk
  • apoorvk
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.
Mimi_x3
  • Mimi_x3
yeah, and i dont think your answer is right..
Mimi_x3
  • Mimi_x3
the binomial expansion: \[(a+b)^n = \binom{n}{r} *(a)^{n-r} * (b)^{n}\]
apoorvk
  • apoorvk
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..
Mimi_x3
  • Mimi_x3
nope..
Mimi_x3
  • Mimi_x3
\[\binom{6}{3} *(0.61)^3*(0.39)+\binom{4}{1} *(0.80)^1*(0.20)^3\] apparently not right
apoorvk
  • apoorvk
you obviously need to multiply the two probabilities, not add them - think about it.
Mimi_x3
  • Mimi_x3
still not right
Mimi_x3
  • Mimi_x3
close though
Mimi_x3
  • Mimi_x3
the answer is \(0.0060\)
apoorvk
  • apoorvk
and what do you seem to be getting?
Mimi_x3
  • Mimi_x3
\(0.0453\) man i hate probability; forget it
apoorvk
  • apoorvk
I am starting to hate it too -...-
Mimi_x3
  • Mimi_x3
lol its horrible; too confusing -_-

Looking for something else?

Not the answer you are looking for? Search for more explanations.