Math help

[FlipinAsian]

People always tell me that I'm very bright. I guess I am, but I learn at extremely slow paces. I'm surprised I made it this far in school, without getting held back. But once I understand something, it's smooth sailing from there.

(Yep, didn't learn how to spell "blue" until I was in second grade, but look at me now! Spelling like a conflagration that can't be tamed.)

Anyway.. here is the math problem.


Cars driving north to an intersection experience a rotating traffice light signal cycle with 55 seconds of red followed by 25 seconds of green. If someone drives randomly to this interestion from the north four times in one week, find the probabilities that the light will be green exactly one time. Also, find the probability it will be green exactly two times.


Call me an idiot or dummy, but I could not find a way to solve this. On my paper, I put "Cannot be determined." I'm pretty much on the risk of flunking because I can't grasp probabilities or statistics!


I look like a valet parker!
I'm reaching out to ya' man! You gotta help me!
HEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPP PPPPPPPPPP!!!!


:chain: Only ONE more semester! C'mon Flippy, you can pull this off!

[IsZi]

It's figurable... gimme a bit, and I think I can get it for you.

Actually, if you typed the problem EXACTLY the way it is written on the test, you are right in saying that it could be unsolvable. We are given values of the red and green lights in on cycle for "Cars driving north to an intersection" (meaning cars travelling northbound), but are asked to solve for "someone (driving) randomly to this intersection from the north" (which would mean he's headed southbound).

But, skipping the possible error in typing, here's how the problem can be solved. Hope you can follow my line of thought here, assuming I am right.

We are given that a traffic light has a red light for 55s, followed by a green light for 25s, so put that up in our mental chalkboard:

red: 55s
green: 25s

Add that together and you get:

total cycle: 80s

So, since you have a green light for 25s out of an 80s cycle, your probability of getting a green light at a random time in any one given cycle would be:

probability of green/cycle: 25:80

Now, we have a car coming to this intersection at random 4 times a week, looking for a green light just one time. To solve for this, we must start by finding the probability of him hitting a green light once if he just tried it one time in a week. First, for this, we must find the amount of seconds in a week:

60 s/m * 60 m/hr * 24 hr/dy * 7 dy/wk = 604800 s/wk

Next, divide by the number of seconds in one traffic cycle, to find the amount of cycles per week:

604800 s/week / 80 s/cycle = 7560 cycles/wk

Now, multiply the time for a green light in a cycle by the number of cycles per week, and:

25 s green/cycle * 7560 cycles/wk = 189000 s green/wk

Put that next to your amount of seconds per week, and you have your probability of hitting a green light in one stop per week:

189000:604800

[edit]
Crap... totally fragged myself from after here, but this should be a good start. Somone pick up?

[FlipinAsian]

:?

Yes, I was under the assumption that the cycle-rates for north and southbound were not the same. Heh, I guess the shortcut was not the way out this time.

[-=DoW=- Saw_Gunner69]

ahhhhh but wise coby for got one thing...i have yet to see a traffic light with out a yellow on it....what is the cycle tiem of the yellow??? :banger:

[FlipinAsian]

That factor was never stated.
I'm guess we are assuming this was before the yellow light was implemented into the system.

[Panda]

I'd solve it for you, but I'm still working on Ace's problem for me. :mrgreen:

[JL]

I can't even read the problem.