Red Car Conjunction

Pavel314 · science forum addict Joined: 29 Apr 2005 Posts: 78

The other day there was a horrendous traffic jam on the way to work. At one
point, I noticed that there was a red car in front of me and a red car
beside me; my car is red, so this conjunction of three red cars seemed a bit
unusual. I wondered how unusual it was.

To solve this, I walked through three parking lots that lunchtime and came
up with the following percentages of red cars in each lot:

Lot A: 9 red out of 83, 10.8%
Lot B: 10 red out of 117, 8.5%
Lot C: 13 red out of 114, 11.4%

Average of the three samples is 10.3% red, Standard Deviation 1.5%. For ease
of calculation, I'll use 10% red in the following calculations.

How do I interpret this? Or maybe the question should be "How do I ask
this?".

If I ask the odds of three cars at random being red, it seems the answer is
(10%)^3 or 0.1%.

If I ask the odds of two cars next to me being red, the answer is (10%)^2 or
1%.

Given that there were two lanes of traffic headed in the same direction,
three red cars in conjunction would occur in a block of six cars as shown
below.

1...2...3---->
4...5...6---->

If my car is 2, cars 1 and 5 were red that morning. I would have also
counted 1-2-6 red, 1-2-3 red, etc., as a conjunction. Diagonals count, too.

So maybe the probability of getting exactly three cars in the block to be
red is calculated as:

6C3 * (10%)^3 * (1 - 10%) ^ (6 - 3) = 20 * .001 * .729 = 1.46%

While the calculations above are for a static condition, the cars around me
changed continuously, although not very quickly. Cars turned off onto side
streets, other cars joined the throng, one lane would move while the other
stood still, etc. I would estimate that the 3-red conjunction lasted at most
2 minutes of the 90-minute commute, which would be 2.22% of the total time.

Any corrections, comments or clues will be appreciated.

Paul

Michael Zedeler · science forum beginner Joined: 29 Nov 2005 Posts: 17

Pavel314 wrote:

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Paul wrote:
"The other day there was a horrendous traffic jam on the way to work. At
one point, I noticed that there was a red car in front of me and a red
car beside me; my car is red, so this conjunction of three red cars
seemed a bit unusual. I wondered how unusual it was.
To solve this, I walked through three parking lots that lunchtime and
came up with the following percentages of red cars in each lot:
Lot A: 9 red out of 83, 10.8%
Lot B: 10 red out of 117, 8.5%
Lot C: 13 red out of 114, 11.4%
Average of the three samples is 10.3% red, Standard Deviation 1.5%. For
ease of calculation, I'll use 10% red in the following calculations.
How do I interpret this? Or maybe the question should be "How do I ask
this?".
If I ask the odds of three cars at random being red, it seems the answer
is (10%)^3 or 0.1%.
If I ask the odds of two cars next to me being red, the answer is
(10%)^2 or 1%.
Given that there were two lanes of traffic headed in the same direction,
three red cars in conjunction would occur in a block of six cars as
shown below.
1...2...3---->
4...5...6---->
If my car is 2, cars 1 and 5 were red that morning. I would have also
counted 1-2-6 red, 1-2-3 red, etc., as a conjunction. Diagonals count,
too.
So maybe the probability of getting exactly three cars in the block to
be red is calculated as:
6C3 * (10%)^3 * (1 - 10%) ^ (6 - 3) = 20 * .001 * .729 = 1.46%.
______________________________________
Re:
You can't include "your" car in the calculation of red car combinations.
It is red 100% of the time and is, presumably, present in all
combinations you witness.
Therefore, the question to ask is: "What is the probability of two of
the eight cars surrounding my car, in a block of nine cars total, being
red?"
Well, if the probability of any car being red is indeed 10% then the
answer is:

(8C2)(0.1^2)(0.9^6)=14.9%

For 3 of the eight surrounding cars being red the answer is:

(8C3)(0.1^3)((0.9^5)=3.3%

For none of the cars being red:

(8C0)(0.1^0)(0.9^ Cool

= 0.9^8= 43.0%

It is an application of the binomial distribution.

Dan Akers

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Paul wrote:
"Given that there were two lanes of traffic headed in the same
direction, three red cars in conjunction would occur in a block of six
cars as shown below.
1...2...3---->
4...5...6---->
If my car is 2, cars 1 and 5 were red that morning. I would have also
counted 1-2-6 red, 1-2-3 red, etc., as a conjunction. Diagonals count,
too.
So maybe the probability of getting exactly three cars in the block to
be red is calculated as:
6C3 * (10%)^3 * (1 - 10%) ^ (6 - 3) = 20 * .001 * .729 = 1.46%
While the calculations above are for a static condition, the cars around
me changed continuously, although not very quickly. Cars turned off onto
side streets, other cars joined the throng, one lane would move while
the other stood still, etc. I would estimate that the 3-red conjunction
lasted at most 2 minutes of the 90-minute commute, which would be 2.22%
of the total time."
_____________________________________
Re;
Sorry, I didn't really answer your question; I was thinking 3 lanes of
traffic, 9-car block, in my previous post. In any event, you can't
include your car in the calculation. So, for two lanes, the question
would be: "What is the probability that two of the five cars adjoining
mine in a six car block are red?"

The answer would be:
5C2(0.1^2)(0.9^3)= 7.3%

I hope that helps...

Dan Akers

Pavel314 · science forum addict Joined: 29 Apr 2005 Posts: 78

Thanks, Dan. Good point that since my red car is always there, I have to
treat it as a constant.

Paul

"D. Akers" <digikey@webtv.net> wrote in message
news:4489-445102B2-257@storefull-3135.bay.webtv.net...
Paul wrote:
"The other day there was a horrendous traffic jam on the way to work. At
one point, I noticed that there was a red car in front of me and a red
car beside me; my car is red, so this conjunction of three red cars
seemed a bit unusual. I wondered how unusual it was.
To solve this, I walked through three parking lots that lunchtime and
came up with the following percentages of red cars in each lot:
Lot A: 9 red out of 83, 10.8%
Lot B: 10 red out of 117, 8.5%
Lot C: 13 red out of 114, 11.4%
Average of the three samples is 10.3% red, Standard Deviation 1.5%. For
ease of calculation, I'll use 10% red in the following calculations.
How do I interpret this? Or maybe the question should be "How do I ask
this?".
If I ask the odds of three cars at random being red, it seems the answer
is (10%)^3 or 0.1%.
If I ask the odds of two cars next to me being red, the answer is
(10%)^2 or 1%.
Given that there were two lanes of traffic headed in the same direction,
three red cars in conjunction would occur in a block of six cars as
shown below.
1...2...3---->
4...5...6---->
If my car is 2, cars 1 and 5 were red that morning. I would have also
counted 1-2-6 red, 1-2-3 red, etc., as a conjunction. Diagonals count,
too.
So maybe the probability of getting exactly three cars in the block to
be red is calculated as:
6C3 * (10%)^3 * (1 - 10%) ^ (6 - 3) = 20 * .001 * .729 = 1.46%.
______________________________________
Re:
You can't include "your" car in the calculation of red car combinations.
It is red 100% of the time and is, presumably, present in all
combinations you witness.
Therefore, the question to ask is: "What is the probability of two of
the eight cars surrounding my car, in a block of nine cars total, being
red?"
Well, if the probability of any car being red is indeed 10% then the
answer is:

(8C2)(0.1^2)(0.9^6)=14.9%

For 3 of the eight surrounding cars being red the answer is:

(8C3)(0.1^3)((0.9^5)=3.3%

For none of the cars being red:

(8C0)(0.1^0)(0.9^ Cool

= 0.9^8= 43.0%

It is an application of the binomial distribution.

Dan Akers

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