Probability

Hong Kong

Stage 1 - Stage 3

Lesson

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`').

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.

Traffic lights showing green, red and yellow lights are set up on a race track. A race car approaching the lights is equally likely to see a green, red or yellow light.

This is because:

the three lights stay on for the same amount of time.

Athe race car driver is driving at a constant speed.

Bthe lights are the same size.

C

Every $60$60 seconds, a traffic light remains green for $34$34 seconds, yellow for $3$3 seconds and red for $23$23 seconds.

Which outcome is more likely?

Arriving at the traffic light when it is yellow.

AArriving at the traffic light when it is red.

BArriving at the traffic light when it is green.

COn her way to work, Nadia passes through $3$3 sets of such lights. What is the probability that none of the lights are green?

What is the probability that the first two sets of lights are green and red at the third set?