Help on Pre-Calc Homework (Probability)

[Campking]

A(B) = C! / [B! * (C - B)!] * p^B * q^(C - B)
C = Number of chances for it to happen
B = Number you want to happen
p - success prob
q - fail prob
(! = Factorial if you didnt know)

Bionomial Equations are fun. Just write them down clearly.

[Fir3]

a(b) = c! / [b! * (c - b)!] * p^b * q^(c - b)
c = number of chances for it to happen
b = number you want to happen
p - success prob
q - fail prob
(! = factorial if you didnt know)

bionomial equations are fun. Just write them down clearly.

*mindblown* xo

Thanks, we'll see if I can do it.. =/

[iRitch]

So, I need help on this question, if anyone could help and explain that'd be greatly appreciated.

Find the probability of getting two green lights in a row out of five traffic signal lights. Assume red, green, and yellow are equally likely occurences.

(It's mutiple choice, but I'd really appreciate if you could explain it, my teacher already said he wouldn't help ANYONE on this because 3/4's of our class didn't do their assignment.)

A.) 4/81 B.) 80/243 C.) 2/5 D.) 32/243

Thanks Guys!


I know it's kinda stupid to come online to ask help on my homework, but oh well =/

Problem is messed up...no right answer...If the chances of hitting yellow red and green are all the same then the answer would be 1/243...you have a 1/3 chance at every light..and the chances of hitting the second are also 1/3 therefore it would be 1/9, but since there are five lights you have to do 1/3*1/3*1/3*1/3*1/3...this is all i could think of :P sorry

Also if you do 1/3*2/5 you get 2/15. Unfortunately, if you do the 2/15*2/15*2/15*2/15*2/15 you get 32/bignumber. i'm just putting what I have gotten/worked through my mind

[mgrapes]

I think the answer is B), although I'm getting 79/243 rather than 80/243. I don't remember enough probability to be able to tell you how to calculate this using formulas, but I can show you the grad student way (aka Mathematica):

Let red = r, green = g, and yellow = y. We begin by listing all the ways that you could get five lights choosing from the three options each time:

http://www.wolframalpha.com/input/?i...%2Cg%2Cy}%2C5]

Since there's an equal probability of choosing each, the possibilities here are pretty trivial to calculate: 3^5 = 243. We can confirm this by counting the possibilities from above:

http://www.wolframalpha.com/input/?i...2Cg%2Cy}%2C5]]

Now we're only interested in those options which have two greens in a row. We can count them like this (I would have listed them for you first, but WolframAlpha doesn't support the Cases function):

http://www.wolframalpha.com/input/?i...Cg%2Cg%2C___}]

We have 79 cases of two greens in a row out of 243 possibilities, so the answer is 79/243. That said I'm a little uncomfortable with the discrepancy between 79 and 80, so let me think about it a bit more and see if I can come up with a mathematical explanation. But I'm 99% sure that's the right answer

Mike

[Fir3]

Thanks Everyone! I have until Tuesday, so I have some time to figure out the answer! I'll keep checking back to see if anyone else has figured it out. @Mike, Your answer has enough to back it up. I'll check back to see if you figured out why you came up with 79 and not 80..

[iRitch]

I'm about 89% sure its D...

[Joe Fries]

My first guess was B or D without working the problem. This question is worded terribly. You should tell your professor to be more specific. Does he mean the probability of getting any combination with 2 greens in a row? What about a (g,g,y,g,g)? Does that count as 2? This is a terrible question.

[Antidote]

I'm about 89% sure its D...

39% of all statistics are made up on the spot.

[Eclipse]

39% of all statistics are made up on the spot.

Evidence or this statement is false......lol