Help on Pre-Calc Homework (Probability)

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 =/

im in precalc and i have no idea what this means

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Originally Posted by chris_cr33p

im in precalc and i have no idea what this means

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Hahaha, it's actually statistics, but we're being taught this in pre-calc.

So the probability of getting the first green light is 1/3.
Then the chance of getting the second one in a row would also be 1/3
The probability of this is 1/9 (1/3*1/3)
But your chances are greater than that because you have the chance to get 2 in a row 4 times...

...Still thinking...

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Originally Posted by RockonitsAjay

So the probability of getting the first green light is 1/3.
Then the chance of getting the second one in a row would also be 1/3
The probability of this is 1/9 (1/3*1/3)
But your chances are greater than that because you have the chance to get 2 in a row 4 times...

...Still thinking...

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That's kinda how I started out, but I got lost after that... Thanks for trying! :)

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Originally Posted by Fir3

That's kinda how I started out, but I got lost after that... Thanks for trying! :)

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From what I have my best guess would be A 4/81 because it has the 4 (from the four chances to get 2 consecutively) and 81 which is the 9*9

I'm really wishing I can solve this haha
Reminds me of math team from high school. No shame.

EDIT: Never mind I think your chances are greater than 5%

Well I have until Tuesday, so we'll see if someone can come up with the answer, but my guess was A also.

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Originally Posted by Fir3

Well I have until Tuesday, so we'll see if someone can come up with the answer, but my guess was A also.

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Well gl.
Want to help me study for my 5 exams this week?
I'm solving your math problem instead of studying and I'm a Sociology major lol.

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Originally Posted by RockonitsAjay

Well gl.
Want to help me study for my 5 exams this week?
I'm solving your math problem instead of studying and I'm a Sociology major lol.

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If you want to pass any of your exams, I better not.. Haha!

lllllllllllllllol

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.

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Originally Posted by 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.

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*mindblown* xo

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

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Originally Posted by Fir3

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 =/

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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

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

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.. :)

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

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.

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Originally Posted by iRitch

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

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

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Originally Posted by Antidote

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

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