Traffic Light Problems

The important feature of probability questions involving time and physical space is that the sample space is continuous. There is an uncountable infinity of outcomes in the sample space if our focus is on instants of time or on individual points in physical space. This means that probabilities cannot be formed by counting the elements in the sets involved. 

Instead, we measure intervals of time or space and we let the relative size of the interval compared with the measure of the whole space be the probability. Thus, if a light in a room is turned on for 45 seconds in every five minutes, we would say that the probability of a person who enters the room at a random moment encountering the light 'on' is $\frac{45}{5\times60}=0.15$455×60​=0.15.

Probabilities determined in this way behave in the same way as do probabilities in a discrete sample space.

In particular, it remains true that for mutually exclusive events $A$A and $B$B - that is, for non-overlapping intervals in the space - we have $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)$P(A∪B)=P(A)+P(B) (where $A\cup B$A∪B is read as '$A$A or $B$B').

Also, if $A$A and $B$B are independent events, $P\left(A\cap B\right)=P\left(A\right).P\left(B\right)$P(A∩B)=P(A).P(B) (where $A\cap B$A∩B is read as '$A$A and $B$B').

 

Example

The following diagram represents a road network in which at each node exactly one coloured road is available at any moment. In each 3-minute period, the red roads are available for 42 seconds and the green roads are available for 78 seconds. Otherwise, the blue roads are available. Where there is a choice, vehicles turn to the left or the right equally often onto a particular coloured road. They cannot change direction on the same road. What is the probability that a vehicle leaving A will arrive at node B having travelled on three road segments?

There are six paths from A to B that have three steps. With their associated probabilities (deduced from the time intervals) they are:
1.  green $\left(\frac{78}{180}\approx0.43\right)$(78180​≈0.43), green $\left(\frac{78}{180}\approx0.43\right)$(78180​≈0.43), blue $\left(\frac{60}{180}\approx0.33\right)$(60180​≈0.33).
2.  green $\left(\frac{78}{180}\approx0.43\right)$(78180​≈0.43), blue $\left(\frac{30}{180}\approx0.17\right)$(30180​≈0.17), green $\left(\frac{36}{180}=0.2\right)$(36180​=0.2)
3.  red $\left(\frac{42}{180}\approx0.23\right)$(42180​≈0.23), red $\left(\frac{42}{180}\approx0.23\right)$(42180​≈0.23), blue $\left(\frac{30}{180}\approx0.17\right)$(30180​≈0.17).
4.  red $\left(\frac{42}{180}\approx0.23\right)$(42180​≈0.23), blue $\left(\frac{30}{180}\approx0.17\right)$(30180​≈0.17), red $\left(\frac{21}{180}\approx0.12\right)$(21180​≈0.12).
5.  blue $\left(\frac{60}{180}\approx0.33\right)$(60180​≈0.33), green $\left(\frac{78}{180}\approx0.43\right)$(78180​≈0.43), green $\left(\frac{78}{180}\approx0.43\right)$(78180​≈0.43).
6.  blue $\left(\frac{60}{180}\approx0.33\right)$(60180​≈0.33), red $\left(\frac{42}{180}\approx0.23\right)$(42180​≈0.23), red $\left(\frac{42}{180}\approx0.23\right)$(42180​≈0.23).

Multiplying the probabilities along each path, we have:
1.  $\frac{169}{2700}$1692700​
2.  $\frac{13}{900}$13900​
3.  $\frac{49}{5400}$495400​
4.  $\frac{49}{10800}$4910800​
5.  $\frac{169}{2700}$1692700​
6.  $\frac{49}{2700}$492700​

Each path is a single outcome in the experiment. That is, the paths are mutually exclusive. Therefore, we add the probabilities and obtain $\frac{617}{3600}\approx0.17$6173600​≈0.17.

As a reasonableness check on this result, note that there are 13 nodes that can be reached from A in three steps under the given rules. Several of these can only be reached in one way. So, we might expect a probability somewhat larger than $\frac{1}{13}\approx0.077$113​≈0.077.

Worked Examples

QUESTION 1

QUESTION 2