Section 1
What is Probability? The Chance of an Event
What is probability? It is simply the chance of something happening.
Take a basic coin toss. What is the chance of getting a tail?
$P(\text{Tail}) = \frac{1}{2}$ (1 favourable outcome out of 2 total possibilities)
The Fundamental Rule
$E = \text{Event}$$P(E) = \frac{\text{Desire}}{\text{Total } \Omega}$ (What you want divided by the total sample space)
Let's look at the Probability Scale:
$0 = \text{Impossible}$
$1 = \frac{\text{Total}}{\text{Total}} = \text{Certain}$ (e.g., $\frac{15}{15} = 1$)
Two Coins Tossed Case Study
Two coins are tossed. What is the probability of getting a tail and a head?
First, write out the explicit Sample Space to see all 4 outcomes:
$\text{Sample Space} = \{HH, HT, TH, TT\}$
Each individual branch has a chance of happening:
$HH = \frac{1}{4}$, $HT = \frac{1}{4}$, $TH = \frac{1}{4}$, $TT = \frac{1}{4}$
To get a tail and a head, order doesn't matter yet, so we could get $HT$ OR $TH$.
Must add to total! Because it's an OR condition, we add the probabilities.
$\text{Probability} = HT + TH = \frac{1}{4} + \frac{1}{4}$
$= \frac{1}{2}$
Alternative Branch Logic: $\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Section 2
Sequential Sampling: Replaced Marbles
There are 5 blue and 3 green marbles in a bag. Total = 8 marbles.
One marble is taken out, recorded, and then REPLACED. A second marble is then taken out.
Let's build the tree diagram branches: Blue = $\frac{5}{8}$, Green = $\frac{3}{8}$.
(i) Find the probability that both are the same colour.
This means we want (Blue AND Blue) OR (Green AND Green).
$\text{P(BB)} = \frac{5}{8} \times \frac{5}{8} = \frac{25}{64}$
$\text{P(GG)} = \frac{3}{8} \times \frac{3}{8} = \frac{9}{64}$
$\text{Total} = \frac{25}{64} + \frac{9}{64}$
$= \frac{34}{64} = \frac{17}{32}$
(ii) Find the probability that at least one is blue.
"At least one blue" means we accept $BB$, $BG$, or $GB$.
$\text{P(BB)} = \frac{25}{64}$
$\text{P(BG)} = \frac{5}{8} \times \frac{3}{8} = \frac{15}{64}$
$\text{P(GB)} = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}$
$\text{Total} = \frac{25}{64} + \frac{15}{64} + \frac{15}{64}$
$= \frac{55}{64}$
Section 3
Simultaneous Sampling: Without Replacement
There are 6 green marbles and 4 pink marbles in a bag. Total = 10 marbles.
Two marbles are taken out simultaneously.
VITAL LOGIC TRAP!
Simultaneously means NOT REPLACED. The pool shrinks on the second draw!
(i) Find the probability that both are the same colour.
We want (Green AND Green) OR (Pink AND Pink).
$\text{P(GG)} = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$
$\text{P(PP)} = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{4}{15}$
$\text{Total} = \frac{30}{90} + \frac{12}{90} = \frac{42}{90}$
$= \frac{7}{15}$
Alternative Combinations Method: $\frac{^{6}C_{2} \times ^{4}C_{0}}{^{10}C_{2}} + \frac{^{6}C_{0} \times ^{4}C_{2}}{^{10}C_{2}}$
(ii) Find the probability that at least one is pink.
Strategic Shortcut: The opposite of "at least one pink" is "both green". The opposite is a buy!
$\text{P(At least one Pink)} = 1 - \text{P(Both Green)}$
$1 - \text{P(GG)} = 1 - \frac{1}{3}$
$= \frac{2}{3}$
Section 4
Two Dice Results Table Grid
Two dice are thrown and their results are added. Find the probability of getting a total result of 7 or more.
Let's look at the $6 \times 6 = 36$ outcomes grid layout:
$(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$
$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)$
$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)$
$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)$
$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$
$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$
$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)$
$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)$
$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)$
$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$
$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$
Counting the combinations that add up to exactly 7 or more:
There are 21 winning combinations out of 36 total.
$\text{Probability} = \frac{21}{36}$
$= \frac{7}{12}$
Section 5
Exam Challenge: Palindromic Numbers
A palindromic number is one that reads the same backwards as forwards, such as 727 or 38183.
(i) This year, 2002, is a palindromic year. When is the next palindromic year?
$2112$
(ii) How many palindromic years are there from 1000 to 9999 inclusive?
Let's map out the 4 slots: $\underline{\text{Thousands}} \times \underline{\text{Centuries}} \times \underline{\text{Tens}} \times \underline{\text{Units}}$
THE TRAP!
The first number cannot be 0! Choices are $\{1,2,3,4,5,6,7,8,9\}$ ($9$ choices).
The centuries box can be anything from 0 to 9 ($10$ choices).
Because it must read the same backwards, the tens digit MUST match the centuries digit ($1$ forced choice), and the units digit MUST match the thousands digit ($1$ forced choice).
$\text{Total} = \underline{9} \times \underline{10} \times \underline{1} \times \underline{1}$
$= 90 \text{ years}$
(iii) A whole number, greater than 9 and less than 10,000, is selected at random. What is the probability that the number is palindromic?
First, calculate the total pool size: numbers from 10 to 9999 inclusive.
$\text{Total pool} = 10,000 - 10 = 9990 \text{ numbers}$
Now count all palindromes in this range by breaking it down by digit size:
2-digit palindromes (e.g., 11, 22, 33...): $\underline{9} \times \underline{1} = 9$
3-digit palindromes: $\underline{9} \times \underline{10} \times \underline{1} = 90$
4-digit palindromes (from part ii): $\underline{9} \times \underline{10} \times \underline{1} \times \underline{1} = 90$
$\text{Total Palindromes} = 9 + 90 + 90 = 189$
$\text{Probability} = \frac{189}{9990}$
$= \frac{21}{1110} = \frac{7}{370}$
Section 6
Exam Challenge: The Girl/Boy Row Permutations
(i) In how many different ways can eight people be seated in a row?
$8! = 40,320 \text{ ways}$
(ii) Three girls and five boys sit in a row, arranged at random. Find the probability that the three girls are seated together.
Remember the Bubble Method: Bound the 3 girls into 1 single bubble element.
Total objects to arrange now = $1 \text{ bubble} + 5 \text{ boys} = 6 \text{ units}$. They arrange in $6!$ ways.
But the girls can shuffle internal positions inside their bubble in $3!$ ways.
$\text{Favourable Arrangements Together} = 3! \times 6!$
$\text{Total Unrestricted Arrangements} = 8!$
$\text{Probability} = \frac{3! \times 6!}{8!}$
$= \frac{6 \times 720}{40320} = \frac{3}{28}$
(iii) Three girls and $n$ boys sit in a row, arranged at random. If the probability that the three girls are seated together is $\frac{1}{35}$, find the value of $n$.
Total number of people = $n + 3$. Therefore, $\text{Total Arrangements} = (n+3)!$
Using the Bubble Method again: Bound the 3 girls. Remaining elements to move = $1 \text{ bubble} + n \text{ boys} = (n+1) \text{ units}$.
$\text{Favourable Count} = 3! \times (n+1)!$
$\text{Probability Equation} = \frac{3! \times (n+1)!}{(n+3)!} = \frac{1}{35}$
Let's expand the larger factorial at the bottom to cancel terms out:
$\frac{6 \times (n+1)!}{(n+3)(n+2)(n+1)!} = \frac{1}{35}$
$\frac{6}{(n+3)(n+2)} = \frac{1}{35}$
Cross multiply to create your quadratic layout:
$(n+3)(n+2) = 6 \times 35$
$n^2 + 5n + 6 = 210$
$n^2 + 5n - 204 = 0$
Factoring the quadratic equation:
$(n-12)(n+17) = 0$
Solutions: $n = 12$ or $n = -17$
Reject $n = -17$. Count must be a positive integer ($n \in \mathbb{N}$).
$= 12 \text{ boys}$
Section 7
Exam Challenge: Coin Mix Algebraic Variables
A box contains four silver coins, two gold coins, and $x$ copper coins. Two coins are picked at random, and without replacement, from the box.
$\text{Total Coins} = 4 + 2 + x = x + 6$
(i) Write down an expression in $x$ for the probability that the two coins are copper.
$\text{P(First is Copper)} = \frac{x}{x+6}$
$\text{P(Second is Copper)} = \frac{x-1}{x+5}$ (Pool shrank by 1)
$\text{Expression} = \frac{x(x-1)}{(x+6)(x+5)} = \frac{x^2 - x}{x^2 + 11x + 30}$
(ii) If it is known that the probability of picking two copper coins is $\frac{4}{13}$, how many coins are in the box?
Equate your expression to the given fraction and resolve:
$\frac{x(x-1)}{(x+6)(x+5)} = \frac{4}{13}$
$13(x^2 - x) = 4(x^2 + 11x + 30)$
$13x^2 - 13x = 4x^2 + 44x + 120$
$9x^2 - 57x - 120 = 0$
Divide the entire row by 3 to simplify the numbers before factoring:
$3x^2 - 19x - 40 = 0$
Splitting the middle term: $-24x + 5x = -19x$
$3x^2 - 24x + 5x - 40 = 0$
$3x(x-8) + 5(x-8) = 0$
$(x-8)(3x+5) = 0$
Solutions: $x = 8$ or $x = -\frac{5}{3}$
Reject the negative fraction value. Coins must be positive whole integers. Therefore, $x = 8$.
Total coins count calculation:
$\text{Total Coins} = 4 + 2 + 8 = 14 \text{ coins}$
(iii) What is the probability that neither of the two coins picked is copper?
"Neither is copper" means we only pick from the remaining non-copper pool (4 silver + 2 gold = 6 coins).
$\text{Total non-copper pool} = 6 \text{ coins}$
$\text{Total coins in box} = 14 \text{ coins}$
$\text{P(Neither Copper)} = \frac{6}{14} \times \frac{5}{13}$
$= \frac{30}{182} = \frac{15}{91}$
End of lesson
Basic Probability — well done!