Intro To Probability

Probability is a complex and fundamental concept in the field of Applied Mathematics. Probability can be described as the chance or likelihood that a particular event, anything that can or will happen, will occur. It is crucial to understand the modelling of the unknown in order to make important informed decisions.

There are fundamental characteristics of probability that can be denoted:

1. The probability of an event, $P(E)$, can occur between the range: $0 \le P(E) \le 1$. In this situation, $0$ means the event occuring is impossible, whereas $1$ means the event occuring is definite.
   

2. In a given sample space $S$, the probability of all the events occuring, $P(S) = 1$.
   

3. In a given sample space, $S$, if events {$E_1, E_2, E_3, ..., E_n$} are mutually exclusive, then the probability of their union, $\cup$, is the sum of their individual probabilities: $P(\bigcup E_n) = \sum P(E_n)$. Furthermore, if the events are also collectively exhaustive, then $P(S) = \sum P(E_n) = 1$.
   

Therefore, we can also denote what the probability of an event, $A$ is:

For any given event, $A$, within a defined sample space, $S$:

$$\begin{align*} P(A) &= \frac{n(A)}{n(S)} \\ \end{align*}$$

In which,

$$\begin{align*} n(A) &= \text{the number of outcomes of event, } A \\ n(S) &= \text{the total number of possible outcomes in the sample space, } S \\ \end{align*}$$

Types of Probability Events

A sample space is the set of all possible outcomes of a random experiment. As aforementioned, an event is anything that can occur or will happen, essentially a subset of a sample space. This means that it be the entire sample space, or the sample space can be made up of multiple events. However, these events can further be specified into the following types:

1. Independent Events

Two events, $A$ and $B$, are said to be independent if the occurence of one event does not affect the occurence of another. Hence:

$$P(A \cap B) = P(A) \times P(B)$$

and,

$$P(A|B) = P(A)$$

2. Dependent Events

Two events, $A$ and $B$, are said to be dependent if the occurence of one event affects the occurence of another. Hence:

$$P(A \cap B) \neq P(A) \times P(B)$$

and,

$$P(A|B) \neq P(A)$$

3. Collectively Exhaustive Events

Two events, $A$ and $B$, are said to be collectively exhaustive if the occurence of both events make up the entire sample space, Hence:

$$P(A \cup B) = 1$$

4. Mutually Exclusive Events

Two events, $A$ and $B$, are said to be mutually exclusive if the event cannot occur simultaneously. Hence:

$$P(A \cap B) = 0$$

and,

$$P(A \cup B) = P(A) + P(B)$$

Unions and Intersections

1. The union of two events, $A$ and $B$, is the event that either A occurs, B occurs or both occur. It is denoted by the symbol, $\cup$, i.e. $P(A \cup B)$. It follows that:
   

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

1. The intersection of two events, $A$ and $B$, is the event that both A and B occur simultaneously. It is denoted by the symbol, $\cap$, i.e. $P(A \cap B)$. It follows that:
   

$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$

Conditional Probability

It is important to note that the probability, the likelihood or chance of an event occuring, can be conditional, meaning that it is the probability given that another event has already occured.

For any two events, $A$ and $B$, the probability that event $A$ occurs given that event $B$ has already occured can be illustrated as:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

and, as aforementioned for independent events,

$$P(A|B) = P(A)$$

Examples

1. Consider a regular coin with sides, heads and tails. If the coin is flipped once, what is the probability of the coin landing on tails?

Solⁿ:

In this situation, we are given a regular two-sided coin which can either land on heads or tails. Therefore, it is intuitive that there are only two options: heads or tails. However, since we only care about the probability that it lands on tails, one of the two options, then the probability where:

Let $A$ be the event of the coin landing on tails.

$$\begin{align*} P(A) &= \frac{n(A)}{n(S)} \\ P(A) &= \frac{1}{2} \\ \end{align*}$$

2. Consider two regular coins with sides, heads and tails. If both coins are flipped once, what is the probability that both coins will land on tails?

Solⁿ:

Conversely, in this situation, we are given two regular two-sided coins which can either land on heads, $H$, or tails, $T$. Therefore, it is intuitive that there are four outcomes:

$$S = {HH, HT, TH, TT}$$

Hence, the probability that both coins will on tails is,

Let $A$ be the event of both coins landing on tails.

$$\begin{align*} P(A) &= \frac{n(A)}{n(S)} \\ P(A) &= \frac{1}{4} \\ \end{align*}$$

3. Consider a fair regular six-sided die which is rolled twice in a game of monopoly. What is the probability that the player rolls a four and then a five after both turns?

Solⁿ:

In this scenario, it can be inferred that the probability of the event of getting a specific number after rolling a die is independent of the previous roll. Therefore, if we consider each roll:

For $1^\text{st}$ roll:

$$S = {1,2,3,4,5,6}$$

So,

Let $A$ be the event that the die will land on four.

$$\begin{align*} P(A) &= \frac{n(A)}{n(S)} \\ P(A) &= \frac{1}{6} \\ \end{align*}$$

For $2^\text{nd}$ roll:

$$S = {1,2,3,4,5,6}$$

So,

Let $B$ be the event that the die will land on five.

$$\begin{align*} P(A) &= \frac{n(A)}{n(S)} \\ P(A) &= \frac{1}{6} \\ \end{align*}$$

Therefore,

Let $C$ be the event that the player rolls a four and then a five

$$\begin{align*} P(C) &= P(A \cap B) \\ P(C) &= P(A) \times P(B) \\ P(C) &= \frac{1}{6} \times \frac{1}{6} \\ P(C) &= \frac{1}{36} \\ \end{align*}$$

Alternatively, we can consider both rolls within an independent sample space altogether:

$$ S = {11, 12, 13, 14, 15, 16, 21, 22, 23, 24, 25, 26, 31, 32, 33, 34, 35, 36, 41, 42, 43, 44, 45, 46, 51, 52, 53, 54, 55, 56, 61, 62, 63, 64, 65, 66} $$

Hence, we can see that out of the $36$ outcomes, there is only $1$ in which the player rolls a four and then a five. Do not mistake the event of getting a five and then a four, $54$, as a possible outcome. The only possible outcome is $45$.

$$\begin{align*} P(A) &= \frac{n(A)}{n(S)} \\ P(A) &= \frac{1}{36} \\ \end{align*}$$

4. In a factory of machines: there is a 20% chance of red, 30% of blue, and 50% chance of yellow plates being manufactured. When a yellow plate is manufactured, 50% of the times it is broken, 30% of times when a blue is made and 25% of times when a red is made. i. What is the probability that a manufactured plate is red AND not broken? ii. What is the probability that the first 10 plates are broken? iii. From your calculations, what can you infer about the machines at the factory?

Solⁿ:

i. What is the probability that a manufactured plate is red AND not broken?

Let $R$ be the event that the plate is red. Let $B$ be the event that the plate is blue. Let $Y$ be the event that the plate is yellow. Let $B_r$ be the event that the plate is broken.

$$\begin{align*} P(R \cap B_r') &= P(R) \times P(B_r') \\ P(R \cap B_r') &= P(R) \times (1-P(B_r)) \\ P(R \cap B_r') &= 0.2 \times (1-0.25) \\ P(R \cap B_r') &= 0.2 \times (0.75) \\ P(R \cap B_r') &= 0.15 \\ \end{align*}$$

Hence, the probability that a manufactured plate is red and NOT broken is: $0.15$.

ii. What is the probability that the first 10 plates are broken?

$$\begin{align*} P(B_r) &= P(R \cap B_r) + P(Y \cap B_r) + P(B \cap B_r) \\ P(B_r) &= (P(R) \times P(B_r|R)) + (P(Y) \times P(B_r|Y)) + (P(B) \times P(B_r|B)) \\ P(B_r) &= (0.2 \times 0.25 ) + (0.5 \times 0.5) + (0.3 \times 0.3) \\ P(B_r) &= (0.05) + (0.25) + (0.0.9) \\ P(B_r) &= 0.39\\ \end{align*}$$

Therefore, the probability for 10 broken plates would follow:

$$\begin{align*} P(10 B_r) &= (0.39)^{10} \\ P(10 B_r) &= 8.14 \times 10^{-5} \text{ (to 3. s.f.)}\\ \end{align*}$$

Hence, we can conlude the probability that the first 10 plates are broken is approximately: $8.14 \times 10^{-5}$

iii. From your calculations, what can you infer about the machines at the factory?

It can be concluded that the machines at the factory deliver an unsatisafactorily high defect rate at around $39%$ of plates. Therefore, the factory needs to reevalute its machinery to ensure that the defect rate may be lowered. It may also be noted that yellow plates are produced the most often and contribute to the majority of broken plates.

Questions

1. James, Jack and Jonesy's absenteeism rates have been recorded for the past month of January. For Jack, he had been absent 20% of the days; For James, 45% of the days; For Jonesy, 35% of the days.

   i. What is the probability that Jack is absent for an entire fortnight? (Ans: $1.6 \times 10^{-10}$)

   ii. What is the probability that neither are present on a given day? (Ans: $0.0315$)

2. In a girls' sports club, 20% play tennis whilst 30% play volleyball. Of the 20% who play tennis, 80% play volleyball. 50% of the persons in the club play another sport.

   i. What is the probability that a girl plays tennis AND volleyball? (Ans: $0.16$)

   ii. What is the probability that a girl plays BOTH tennis OR volleyball? (Ans: $0.34$)

3. In a classroom, there are 6 students. Assume each student’s birthday is equally likely to fall in any of the 12 months of the year, independently of others.

   i. What is the probability that no two students share a birthday month? (Ans: $0.223$)

   ii. What is the probability that at least two students share a birthday month? (Ans: $0.777$)