Probability

Hong Kong

Stage 4 - Stage 5

Lesson

We've already learnt about sample spaces, which are lists of all the possible outcomes of an experiment. Once we understand how to interpret sample spaces, we can calculate the probability of specific outcomes.

Generally:

$\text{Probability of an outcome }=\frac{\text{number of outcomes for a specific event }}{\text{total number of outcomes }}$Probability of an outcome =number of outcomes for a specific event total number of outcomes

For example, say I wanted to know the probability of rolling a $3$3 on a dice. There are *six* possible outcomes in total (The sample space is {$1,2,3,4,5,6$1,2,3,4,5,6}). However, there is only *one* chance that I can roll a $3$3 because there is only one three on a dice.

So, I could write $P(3)=\frac{1}{6}$`P`(3)=16

If an experiment has equally likely outcomes, it means the chance of any outcome occurring is equal. In other words, there is no bias. Bias is a favouring of one outcome over all others.

When we flip an unbiased coin, it is equally likely that we will get heads or tails. In other words, there is a $50%$50% chance of each outcome. Similarly, when we roll a fair dice, there is an equal $\frac{1}{6}$16 chance of rolling any number between a $1$1 and a $6$6.

Remember!

The probability of all possible outcomes is $1$1.

e.g. When rolling a die...$P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$`P`(1)+`P`(2)+`P`(3)+`P`(4)+`P`(5)+`P`(6)=16+16+16+16+16+16$=$=$1$1

A complement of an event are all outcomes that are NOT the event.

The probabilities of complementary events always sum to 1.

$\text{P(A) + P(A')}=1$P(A) + P(A')=1

We can also use this knowledge to find the probability of a complementary event.

$P(A')=1-P(A)$`P`(`A`′)=1−`P`(`A`)

Handy Facts

Cards and dice are often used in probability, so here are a couple of handy facts to remember:

- There are $52$52 cards in a normal pack of cards.
- There are $13$13 cards in each of the $4$4 suits (hearts, diamonds, clubs and spades).
- A normal dice has $6$6 faces

A number is randomly selected from the following list:

$\left\{1,3,3,6,6,6,8,8,8,8,10,10,10,10,10\right\}${1,3,3,6,6,6,8,8,8,8,10,10,10,10,10}

How many numbers are in this list?

$\editable{}$

What is the probability of selecting a $1$1?

What is the probability of selecting a $3$3?

What is the probability of selecting an $8$8?

Which number is most likely to be selected?

$\editable{}$

From a normal deck of cards, what is the probability of selecting:

a two?

A four?

Not a seven?

A red card?

A fifteen?

A face card?

A biased coin is flipped, with heads and tails as possible outcomes. Calculate $P\left(\text{heads}\right)$`P`(heads) if $P\left(\text{tails}\right)$`P`(tails) is $0.64$0.64.