19.08 Independent events and trees

Let's look at calculating theoretical probabilities for situations with multi-stage trials. Such as tossing a coin three times, rolling two dice or passing through multiple intersections.

Independent events

Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. When tossing an unbiased coin repeatedly the probability of the occurrence of a tail on any individual toss is $0.5$0.5. This probability remains unchanged, regardless of whether there has been a run of heads or tails in previous tosses, since the coin has no memory. So each coin toss is an independent event.  The events of tossing a coin and then rolling a die are independent, because they use completely different objects.  The die is not affected by the coin and vice versa.

If events $A$A and $B$B are independent the following property allows us to calculate the probability of both $A$A and $B$B occurring, in other words $P\left(A\cap B\right)$P(A∩B).

 

Caution: This only applies to independent events. 

Worked example

Example 1 

If two dice are rolled, what is the probability of rolling snake eyes (double ones)?

Method 1:

Think: We can set out the sample space of rolling two dice using a table.

Do: The sample space is shown in the following table:

	$1$1	$2$2	$3$3	$4$4	$5$5	$6$6
$1$1	$1,1$1,1	$1,2$1,2	$1,3$1,3	$1,4$1,4	$1,5$1,5	$1,6$1,6
$2$2	$2,1$2,1	$2,2$2,2	$2,3$2,3	$2,4$2,4	$2,5$2,5	$2,6$2,6
$3$3	$3,1$3,1	$3,2$3,2	$3,3$3,3	$3,4$3,4	$3,5$3,5	$3,6$3,6
$4$4	$4,1$4,1	$4,2$4,2	$4,3$4,3	$4,4$4,4	$4,5$4,5	$4,6$4,6
$5$5	$5,1$5,1	$5,2$5,2	$5,3$5,3	$5,4$5,4	$5,5$5,5	$5,6$5,6
$6$6	$6,1$6,1	$6,2$6,2	$6,3$6,3	$6,4$6,4	$6,5$6,5	$6,6$6,6

We can see that there are $36$36 pairs of outcomes, and only one of them is a double $1$1. Therefore the probability of rolling double one is: $P(\text{Double Ones})=\frac{1}{36}$P(Double Ones)=136​.

Method 2:

Think: The two rolls are independent events, so we can multiply the probability of rolling a one on each die.

Do: 

$P\left(\text{Double Ones}\right)$P(Double Ones)	$=$=	$\frac{1}{6}\times\frac{1}{6}$16​×16​
	$=$=	$\frac{1}{36}$136​

 

Example 2

If one ball is selected from bag 1 and one from bag 2 above, what is the probability that both selected balls are red?

The two selections are independent events, so we can multiply the probability of selecting a red from each bag.

$P\left(\text{Red, Red}\right)$P(Red, Red)	$=$=	$\frac{1}{6}\times\frac{1}{2}$16​×12​
	$=$=	$\frac{1}{12}$112​

 

Tree diagrams

Tree diagrams can be very useful to display the outcomes in a multi-stage events. Particularly if the number of options and stages is low. We can summarise the information by putting the probability of each stage on the branch and the outcome at the end of the branch. Probabilities of each outcome can be found by multiplying along a branch.

Worked example

Example 3

The following probability tree diagram shows the outcomes of playing two games of tennis where the probability of winning is $\frac{3}{10}$310​ and the probability of losing is $\frac{7}{10}$710​.  (The probabilities can be written as fractions, decimals or percentages).

(a) Find the probability of the player winning both games.

$P\left(\text{WW}\right)$P(WW)	$=$=	$P\left(W\right)\times P\left(W\right)$P(W)×P(W)
	$=$=	$0.3\times0.3$0.3×0.3
	$=$=	$0.09$0.09

(b) Confirm the sum of the probabilities of all the outcomes is equal to $1$1.

$P\left(\text{WW or WL or LW or LL}\right)$P(WW or WL or LW or LL)	$=$=	$P\left(WW\right)+P\left(WL\right)+P\left(LW\right)+P\left(LL\right)$P(WW)+P(WL)+P(LW)+P(LL)
	$=$=	$0.09+0.21+0.21+0.49$0.09+0.21+0.21+0.49
	$=$=	$1$1

(c) Find the probability of the player winning only one game.

$P\left(\text{WL or LW}\right)$P(WL or LW)	$=$=	$P\left(WL\right)+P\left(LW\right)$P(WL)+P(LW)
	$=$=	$0.3\times0.7+0.7\times0.3$0.3×0.7+0.7×0.3
	$=$=	$0.42$0.42

(d) Find the probability of the player winning at least one game.

$P\left(\text{WW or WL or LW}\right)$P(WW or WL or LW)	$=$=	$P\left(WW\right)+P\left(WL\right)+P\left(LW\right)$P(WW)+P(WL)+P(LW)
	$=$=	$0.3\times0.3+0.3\times0.7+0.7\times0.3$0.3×0.3+0.3×0.7+0.7×0.3
	$=$=	$0.51$0.51

Or alternatively, use the complementary event of losing both games and calculate:

$P\left(\text{At least 1 win}\right)$P(At least 1 win)	$=$=	$1-P\left(LL\right)$1−P(LL)
	$=$=	$1-0.7\times0.7$1−0.7×0.7
	$=$=	$0.51$0.51

Reflect: Would these events be independent in real life? Might winning the first game impact the probability of winning the second game? 

Things to note from using probability trees to calculate probabilities of multiple trials:  

• Multiply along the branches to calculate the probability of individual outcomes
• Add down the list of outcomes to calculate the probability of multiple options. Since each outcome is mutually exclusive.
• The sum of probabilities from different branches stemming from a point should be $1$1 
• The sum of all the outcomes should be $1$1

Practice questions

Question 1

QUestion 2

QUestion 3

Question 4