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10.05 Solving problems with probability


Probability calculations

Many situations in probability can be organised into Venn diagrams, arrays, and tree diagrams to organise the information, determine the size of different groups and do calculations.

The formulas and notation encountered can also help us to organise the information given in a question and calculate different probabilities.

Worked example

There are $124$124 students at a school, $74$74 of them attended the school sports carnival, of which $34$34 were primary students and $40$40 were secondary students. There are a total of $80$80 primary students at school. What is the probability a selected student attended the sports carnival given that they are a secondary student?

Think: If selecting a secondary school student is the event $S$S, and selecting someone that attended the sports carnival is $A$A. We are looking for $P\left(A|S\right)$P(A|S). To find this information we can create and fill in a table that includes a "Total" column and a "Total" row.

Do: Create a table and fill in the values given by the question.

  Primary Secondary Total
Attended $34$34 $40$40 $74$74
Didn't attend
Total $80$80 $124$124

In the first and the last columns we have one piece of information missing, so we can find the values to go into those cells by using subtraction. The number of primary students who attended $\left(34\right)$(34) plus the number of primary students who didn't attend $\left(\text{blank}\right)$(blank) will be equal to the total number of primary students $\left(80\right)$(80), so

$\text{Number of primary student who didn't attend}=80-34=46$Number of primary student who didn't attend=8034=46.

Similarly, looking at the "Total" column,

$\text{Number of students who didn't attend}=124-74=50$Number of students who didn't attend=12474=50.

We write these values in the table:

  Primary Secondary Total
Attended $34$34 $40$40 $74$74
Didn't attend $46$46 $50$50
Total $80$80 $124$124

Now we can use the numbers in the rows in a similar way to find the last two values:

$\text{Number of secondary students who didn't attend}=50-46=4$Number of secondary students who didn't attend=5046=4,

$\text{Number of secondary students}=124-80=44$Number of secondary students=12480=44.

Here is the completed table:

  Primary Secondary Total
Attended $34$34 $40$40 $74$74
Didn't attend $46$46 $4$4 $50$50
Total $80$80 $44$44 $124$124

To find the conditional probability we can use the formula $P\left(A|S\right)=\frac{\text{number in }\left(A\cap S\right)}{\text{number in }\left(S\right)}$P(A|S)=number in (AS)number in (S). The notation $\left(A\cap S\right)$(AS) is the students that are both a secondary student and attended the sports carnival, which is $40$40. Similarly,  $\text{number in }\left(S\right)$number in (S) is the number of secondary students, which is $44$44.

The probability $P\left(A|S\right)$P(A|S) is $\frac{40}{44}$4044, or simplified is, $\frac{10}{11}$1011

Practice questions

Question 1

Given that $P\left(A\cap B\right)=$P(AB)=$0.2$0.2 and $P\left(A\cap B'\right)=$P(AB)=$0.3$0.3.

  1. What is the value of $P\left(A\right)$P(A)?

  2. What is the value of $P\left(B\right)$P(B), given that the events are independent?

  3. Given that $A$A and $B$B are independent find $P\left(A\cup B'\right)$P(AB).

Question 2

Consider the following probability Venn Diagram:

  1. Find $P$P$($($A$A$|$|$B$B$)$). Express your answer as a fully reduced fraction.




Describe the results of two- and three-step chance experiments, both with and without replacements, assign probabilities to outcomes and determine probabilities of events. Investigate the concept of independence.


Use the language of ‘if ....then, ‘given’, ‘of’, ‘knowing that’ to investigate conditional statements and identify common mistakes in interpreting such language.

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