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.
+33661 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!
.
+33661 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
.
+118677 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 $${\frac{{\mathtt{32}}}{{\mathtt{243}}}} = {\mathtt{0.131\: \!687\: \!242\: \!798\: \!353\: \!9}}$$
+118677 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?
+33661 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!
.
