Probability

Hong Kong

Stage 4 - Stage 5

Lesson

Life is full of different events. You may walk or drive to school, you may take your lunch or buy it from the canteen. In probability, we break events up into two groups: *independent* and *dependent* events.

Independent events are events that are not affected by any other event. For example, flipping a coin multiple times is an independent event because each flip is an isolated event. You have an even chance of getting a head or a tail each time and this is not affected by what you flip before.

In contrast, dependent events are those that are affected by previous events. For example, if you have a bag full of different coloured marbles, the probability of selecting a particular colour will change depending on the colour of the marble that was previously selected and taken out.

Mutually exclusive events are events that can't occur at the same time. For example, you can't turn left and right at the same time!

Mutually inclusive events are not as common as mutually exclusive events. Basically, it means that one event cannot occur without the other.

The best way to visualise mutually inclusive/exclusive is in a Venn diagram. The Venn diagram below is showing two exclusive events (A can happen or B can happen) and one inclusive event (the green part- A and B).

Let's look at a more concrete example with numbers. Some numbers are even ($2,4,20$2,4,20 etc), which can be written in the dark green circle. Some numbers are multiples of three ($3,9,15$3,9,15 etc) which can be written in the light green circle. These are mutually exclusive events. What if I asked you for an even that is *also* a multiple of three? Now we need to think of numbers that are mutually inclusive (ie. satisfy both conditions) such as $6,12,18$6,12,18 and so on. These are written in the space where the circles overlap.

Conditional probability is concerned with multiple events and is similar to mutually inclusive events. The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. In other words, we calculate the probability of B, given that A has occurred.

Careful!

A lot of conditional probability questions involve replacement:

- "
*With replacement*" means the two events remain independent (ie. the probabilities don't change) - "
*Without replacement*" means that the two events are dependent (the chances change depending on the previous selection)

Tree diagrams are often used to display conditional events and make our probability calculations easier.

Let's look at an example of a conditional event that can be displayed in a tree diagram.

In a netball game, there are two people who can shoot - the goal shooter (GS) and the goal attack (GA). Let's say the GS in a team attempts the shots $60%$60% of the time and scores $70%$70% of the goals that they attempt, while the GA attempts to shoot $40%$40% of the time and scores $60%$60% of the goals that they attempt.

The blue lines represent the "branches" and we write each possible outcome at the end of each branch (in this case, who takes the shot and whether the goal is scored). The probability of each outcome is written along the branch.

So what is the probability of the GA scoring?

To work out this conditional probability, we multiply along the branches, considering the probability of the goal attack attempting the shot (GA) and then scoring (goal).

So, $\text{P(GA Score)}=0.4\times0.6$P(GA Score)=0.4×0.6$=$=$0.24$0.24. This means that the goal attack will successfully score $24%$24% of the time in a match.

What if we want to know the probability of either person scoring a goal?

Well, as you can see in the diagram below, there are two possible ways this combination can occur: 1) the goal shooter can score or 2) the goal attack can score.

$P(goal)$P(goal) |
$=$= | $0.6\times0.7+0.4\times0.6$0.6×0.7+0.4×0.6 |

$=$= | $0.42+0.24$0.42+0.24 | |

$=$= | $0.66$0.66 |

That means, the team scores $66%$66% of the time. If you do the same process for the probability of the team missing, you'll see that it's $34%$34%. So we know that we have added everything up correctly because scoring and missing are complementary events and $P(goal)+P(miss)=0.66+0.34$`P`(`g``o``a``l`)+`P`(`m``i``s``s`)=0.66+0.34$=$=$1$1.

Remember!

- The probabilities of all your outcomes should add up to $1$1.
- Multiply the probabilities along the branches.

There are four cards marked with the numbers $2$2, $5$5, $8$8, and $9$9. They are put in a box. Two cards are selected at random one after the other without replacement to form a two-digit number.

Draw a tree diagram to illustrate all the possible outcomes.

How many different two-digit numbers can be formed.

What is the probability of obtaining a number less than $59$59?

What is the probability of obtaining an odd number?

What is the probability of obtaining an even number?

What is the probability of obtaining a number greater than $90$90?

What is the probability that the number formed is divisible by 5?

A pile of playing cards has $4$4 diamonds and $3$3 hearts. A second pile has $2$2 diamonds and $5$5 hearts. One card is selected at random from the first pile, then the second.

Construct a tree diagram of this situation with the correct probability on each branch.

What is the probability of selecting two hearts ?

The Venn diagram shown shows the number of students in a school playing the sports of Rugby League, Rugby Union, both or neither.

What is the probability that a student selected at random plays Rugby League?

What is the probability that a student selected at random plays Rugby Union?

What is the probability that a student selected at random plays Rugby League only?

What is the probability that a student selected at random does not play Rugby League?

What is the probability that a student selected randomly does not play Rugby Union?

What is the probability that a student chosen at random plays Rugby Union only?