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.

Question

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.

mimi_x3 · Answer

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 · Answer

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 · Answer

This is called Binomial Probability related to the Binomial Expansion..

apoorvk · Answer

So your expression should be:

$$\large (0.61)^3\times (0.39)^3 \times (0.80) \times (0.20)^3$$

apoorvk · Answer

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 · Answer

yeah, and i dont think your answer is right..

mimi_x3 · Answer

the binomial expansion:
$$(a+b)^n = \binom{n}{r}
 *(a)^{n-r} * (b)^{n}$$

apoorvk · Answer

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 · Answer

nope..

mimi_x3 · Answer

$$\binom{6}{3}  *(0.61)^3*(0.39)+\binom{4}{1}  *(0.80)^1*(0.20)^3$$
apparently not right

apoorvk · Answer

you obviously need to multiply the two probabilities, not add them - think about it.

mimi_x3 · Answer

still not right

mimi_x3 · Answer

close though

mimi_x3 · Answer

the answer is $0.0060$

apoorvk · Answer

and what do you seem to be getting?

mimi_x3 · Answer

$0.0453$ 
man i hate probability; forget it

apoorvk · Answer

I am starting to hate it too -...-

mimi_x3 · Answer

lol its horrible; too confusing -_-