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# Probability

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Select the probability that you will encounter two red lights in a row out of five traffic signal lights. Assume red, green, and yellow are equally likely occurrences.

Apr 27, 2015

#5
+5

I interpreted it as "at least two red lights in a row".

When I tackled the problem first I came up with some reasoning that produced a value of 0.309.  However, when I checked with a Monte-Carlo simulation I consistently got higher values close to 0.325.  When I repeated the problem with just 4 lights I got consistent answers for both methods!  However, I tend to believe my Monte-Carlo approach, because it was very simple (see my explanation below), so I modified my "reasoning" to give the result I came to above, matching the MC approach (there was no real "reasoning" involved, I just adjusted it to get 0.325).

Monte-Carlo approach.

1. I randomly selected 5 numbers between 0 and 1 to represent the signals of the five traffic lights.  If the value was less than 1/3 I designated this as a red light.

2. I then went from traffic light to traffic light in order to see if there were two consecutive lights that were red.  If there were I incremented a counter (c, say) If there weren't I left the counter unchanged (the counter started at 0).

3. I only incremented the counter after the first two consecutive red lights and ignored any further consecutive red lights, because I was looking for "at least two".  (So, for example, sequences like "red, red, red, green, yellow" and "red, red, green, red, red" only counted once not twice.)

4. I repeated that process thousands of times, n times, say (a hundred thousand times, in fact).

5. I then divided the value of the counter by the number of times I did it to get the probability (= c/n).

Needless to say, I wrote a simple program in Mathcad to do this, and ran it several times, always getting a value close to 0.325.

There is no guarantee my MC calculation is right though!

.

Apr 29, 2015

#1
0

Ummmmmm...

1/4

Apr 27, 2015
#2
+5

A Monte-Carlo simulation suggests the probability is:

(1 + 2/3 + 2/3 + (4/3)*(2/3)2)*(1/3)2 = 79/243 ≈ 0.325

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Apr 27, 2015
#3
+5

I am not familiar with the monte-carlo simulations.

And like many of these questions this is open to different interpretations.

Note: this answer has been changed so has my interpretation :)

I am going to interprete it to mean - Exactly 2 red lights in a row and none of the others are red

now they can be

1&2    or    2&3    or    3&4    or     4&5     So that is 4 ways

the number of ways that the 2 reds can come first is    1*1*2*2*2 = 8

So altogether there are    8*4 = 32 ways the you can get exactly 2 red lights in a rowand no other red lights

Now the total number of possibilities is    $${{\mathtt{3}}}^{{\mathtt{5}}} = {\mathtt{243}}$$

So I think that the answer is     $${\frac{{\mathtt{32}}}{{\mathtt{243}}}} = {\mathtt{0.131\: \!687\: \!242\: \!798\: \!353\: \!9}}$$

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Apr 29, 2015
#4
+5

now my original interpretation was 2 red lights in a row and the others can be anything, including red.

I am going to have another go at this one.

If exactly 2 red ones there is  32 posibilities.

If there are 3 red ones then it gets harder  Mmm

I don't know - I'll think about it later Alan, what interpretation have you answered?

Apr 29, 2015
#5
+5

I interpreted it as "at least two red lights in a row".

When I tackled the problem first I came up with some reasoning that produced a value of 0.309.  However, when I checked with a Monte-Carlo simulation I consistently got higher values close to 0.325.  When I repeated the problem with just 4 lights I got consistent answers for both methods!  However, I tend to believe my Monte-Carlo approach, because it was very simple (see my explanation below), so I modified my "reasoning" to give the result I came to above, matching the MC approach (there was no real "reasoning" involved, I just adjusted it to get 0.325).

Monte-Carlo approach.

1. I randomly selected 5 numbers between 0 and 1 to represent the signals of the five traffic lights.  If the value was less than 1/3 I designated this as a red light.

2. I then went from traffic light to traffic light in order to see if there were two consecutive lights that were red.  If there were I incremented a counter (c, say) If there weren't I left the counter unchanged (the counter started at 0).

3. I only incremented the counter after the first two consecutive red lights and ignored any further consecutive red lights, because I was looking for "at least two".  (So, for example, sequences like "red, red, red, green, yellow" and "red, red, green, red, red" only counted once not twice.)

4. I repeated that process thousands of times, n times, say (a hundred thousand times, in fact).

5. I then divided the value of the counter by the number of times I did it to get the probability (= c/n).

Needless to say, I wrote a simple program in Mathcad to do this, and ran it several times, always getting a value close to 0.325.

There is no guarantee my MC calculation is right though!

.

Alan Apr 29, 2015
#6
0

Thanks Alan,

Another  for me to think on. Apr 29, 2015