7.08 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)?
(Note: It does not matter if two separate dice were rolled or a single die was rolled twice)
The two rolls are independent events, so we can multiply the probability of rolling a one on each die.
| $P\left(\text{Double Ones}\right)$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
A fair coin is tossed 600 times.
If it lands on heads 293 times, what is the probability that the next coin toss will land on heads?
If it lands on heads once, what is the probability that the next coin toss will land on heads?
Is the outcome of the next coin toss independent of or dependent on the outcomes of previous coin tosses?
Dependent
AIndependent
B
QUestion 2
A coin is tossed twice.
Construct a tree diagram to identify the sample space of tossing a coin twice.
Use the tree diagram to find the probability of getting two tails.
Use the tree diagram to find the probability of getting at least one tail.
QUestion 3
Each school day, Neil either rides his bike to school or walks. There is a $70%$70% chance Neil will ride his bike.
Draw a weighted probability tree diagram showing Neil’s choices for three consecutive school days
What is the probability that on the fourth day Neil walks to school?
What is the probability that Neil walks to school three days in a row?
Question 4
Robert has found that when playing chess against the computer, he wins $\frac{1}{2}$12 of the time.
What is the probability that he wins two games in a row?
What is the probability that he wins three games in a row?
What is the probability that he wins at least one of three games?
Robert wants to figure out which he has a better chance of: winning at least one of two games or winning at least one of three games. Complete his working.
$\text{P(at least one of two games)}$P(at least one of two games) $=$= $1-\text{P(winning neither)}$1−P(winning neither) $=$= $\editable{}$
$\text{P(at least one of three games)}$P(at least one of three games) $=$= $\editable{}$