can someone please help explain 4-8 to me? how did the answer get (.5)^10 from the conditional probability?

http://members.shaw.ca/ThePais/1121HWb270.htm

please some help. i would really appreciate i t.

According to Poisson's law, the probably that exactly ten cars arrive within the same hour is :

(rate) ^ ten * exp ( - rate) / factorial (ten)

10^10 * exp ( -10) / 10!

~ 0.12511 ( ~ 12.5%)

Now, given that 10 cars have arrived, you must compute the probability that none shows violation.

Since this probablilty is 0.5 (50%) for a single car, the probability that it happens for 10 cars in a row is : 0.5 ^ 10 = 1/1024 ~ 0.000977 (~0.1 %).

Just multiply the two probabilities for the first answer : ~ 0.12511*0.000977 ~ 0.000122 ( 0.0122%)



Repeat the same for a given y :

probability for y cars to arrive = (rate)^y * exp (-rate) / factorial (y) = 10^y * exp(-10) / (y!)

Then, among those y cars, use binomial law to find the probability that exactly 10 have no violations :
(10 among y) * (0.5 ^ 10) * (0.5 ^ (y-10))

which turns out to be :
(10 among y) * 0.5 ^ y

Now multiply both probabiliies, and simplify using the fact that :
(0.5 ^ y) * (10 ^ y) = (0.5*10) ^ y = 5 ^ y

and that :
( 10 among y ) = y ! / ( (10 !) * ( y - 10) ! )


The question should rather be : "that 10 no violation cars" instead of three.

This probability is the sum of the following probabilities :
- exactly 10 cars arrives and, among them, exactly 10 have no violation (we already have calculated that probability at question a)
- exactly 11 cars arrives and, among them, exactly 10 have no violation (use question b with y = 11)
- exactly 12 cars arrives and, among them, exactly 10 have no violation (use question b with y = 12)
- exactly 13 cars arrives and, among them, exactly 10 have no violation
...

Sum from y = 10 to infinity, as the hint suggests.

i'm just a little confused here because the solution says

P(no violations | X = 10)

so isn't that saying given the probability that exactly 10 cars arrive within the same hour, there are no violations.

so can't P(no violations | X = 10) be written as:

P(no violations INTERSECTION X=10) / P(X = 10) ???

This post was edited by iGotThatFiyah on Apr 27 2014 11:24pm

Correct. Generally speaking, p ( A | B ) = p ( A inter B ) / p (B) for any two events A and B (with p(B) =/= 0).

But notice that, in the problem, you are able to find p ( A | B ) directly, and then you can plug in the result to find p ( A inter B ), which is what you're precisely asked for :

p ( A inter B ) = p ( A | B ) * p (B)

I hope it helps, feel free to pm back !