# Probability

Contents

## Summary

• $Probability\quad =\quad \frac { favourable\quad outcomes }{ possible\quad outcomes }$
• Two events that have nothing in common are known as mutually exclusive events $P(A)\quad +\quad P(B)\quad =\quad P(A\quad \cap \quad B)\quad =\quad 0$
• Two events that have elements in common are known as non-mutually exclusive events $P(C\quad \cup \quad D)\quad =\quad P(C)\quad +\quad P(D)\quad -\quad P(C\quad \cap \quad D)$
• If two events are independent event meaning they both do not depend on one another $P(A\quad \cap \quad B)\quad =\quad P(A)\quad \times \quad P(B)$
• If two events are dependent events they both are $P(\frac { A }{ B } )\quad =\quad \frac { P(A\quad \cap \quad B) }{ P(B) }$ (conditional probability)

It is a measure of the likelihood that an event will happen. Meaning, imagine that you roll a dice the possible outcomes of rolling a dice would be {1, 2, 3, 4, 5, 6} that is a set of outcomes, which describe what outcomes correspond to the “event” happening. For example, “rolling a multiple of 3” corresponds to the set of outcomes {3, 6}. We can thus say that the probability of rolling a multiple of 3 is the favourable outcomes divided by the possible outcomes.

$Probability\quad =\quad \frac { favourable\quad outcomes }{ possible\quad outcomes }$

So in this case we can say that the favourable outcome is {3, 6} since that’s the event we are looking for. however, the possible outcomes would be {1, 2, 3, 4, 5, 6} which would be the total outcomes we can get.

Similarly if we toss a coin twice what is the likelihood that we get tails both time.

By using the formula above we can say that the favourable outcome in this case would be:

1 = TT

But the possible outcome would be:

4 = TT, TH, HT, HH

Thus, the probability would be $\frac { 1 }{ 4 }$.

#### Understanding probability through a set notation

We can also study probability using the set notations. A set is a collection of different elements.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

• Empty set

An empty set is an imaginary set which has no elements

$\varphi$ = { } if there are no elements it is known as an empty set.

Note: $\varphi$ is called Phi

• Subset

Is the set in which all the elements are taken from a bigger set.

E.g

A = {2, 4, 7}

”A” is a subset of S as all the elements in ”A” are also present in S.

It can be written as:

$A\quad \subset \quad S$

• Improper and proper subset

Every set has two improper subsets:

1. An empty set
2. And the set itself

E.g

S = {1, 2, 3, 4}

C = { }

B = {1, 2, 3, 4}

C and B are both subsets of S, C is a null set and B is the set itself

• Proper subset

A proper subset is a set which has only a few elements of the main set.

E.g

A = {1, 2, 3, 4}

”A” here is the main subset and both B and C have few elements of A in their set.they are thus, known as a propers subset of A.

B = {1}

C = {1, 2}

#### Sample Space

All possible outcomes of a random experiment

E.g

S = {1, 2, 3, 4, 5, 6} when you roll a dice

S = {H, T} when you toss a coin

S = {HH, HT, TH, TT} when you toss 2 coins

In order to find A’ (A compliment) we subtract set A from the universal set to get A’ and the same is done with set B to find B complement:

A’ = U – A

B’ = U – B

NOTE:

P(A) + P(A’) = 1

Events

Any subset of a sample space is known as an event.

• Mutually Exclusive Events

Two events that have nothing in common are known as mutually exclusive events or disjoint event. In other words if the intersection of two events is none then it’s mutually exclusive.

E.g

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

A = {1, 5}

B = {3, 6}

The venn diagram below shows mutually exclusive events.

However, keep in mind that B alone is not the complement of A.

Complement of A = {2, 3, 4, 6}

So we can write this as:

$A\quad \cap \quad B\quad =\quad$ { }

i.e $P(A)\quad +\quad P(B)\quad =\quad P(A\quad \cap \quad B)\quad =\quad 0$ (known as the addition law of probability for mutually exclusive events)

• Non – Mutually Exclusive Events

If both events have something in common:

E.g

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

C = {3, 4, 5}

D = {4, 5, 6}

Thus it can be written as:

$C\quad \cap \quad D\quad \neq \quad$ { }

$P(C\quad \cap \quad D)\quad =\quad \left\{ 4,\quad 5 \right\}$

$P(C\quad \cup \quad D)\quad =\quad P(C)\quad +\quad P(D)\quad -\quad P(C\quad \cap \quad D)$ (known as the addition law of probability for non-mutually exclusive events)

#### Multiplication law of probability

If two events are independent event meaning they both do not depend on one another.

A and B then the probability of A and B is equal to the probability A and probability B

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

E.g

If there are two sweets in a bag and 7 red coloured and 11 green coloured they are considered independent if the sweet drawn is replaced after pulling out, meaning put back inside the bag. In that case its tree diagram will look like the following:

However if the sweet drawn is not replaced, meaning it is not put back in the bag, in that case the tree diagram will look like the following:

#### Conditional probability

Is the probability of one event, given that another event has already occured.

If two events are dependent events they both are:

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

If both the events are independent then we write it as:

$P(\frac { A }{ B } )\quad =\quad \frac { P(A)P(B) }{ P(B) } \quad =\quad P(A)$

E.g

FemaleMaleTotal
Maths41418
Economics174158
Science42529
Arts281139
Total5391144

a) A student is selected at random from all of the students. Find the probability that

• Student is female

$P(Female)\quad =\quad \frac { 53 }{ 144 }$

• Student is studying arts

$P(Art)\quad =\quad \frac { 39 }{ 144 }$

b) the student is a female and studying arts

$P(F\quad \cap \quad A)\quad =\quad \frac { 28 }{ 144 }$

c) state whether F and A are independent

$P(F\quad \cap \quad A)\quad =\quad P(F)P(A)$

$\frac { 28 }{ 144 } \quad \neq \quad \frac { 53 }{ 144 } \quad \frac { 39 }{ 144 }$

Since, they are not equal we can say that they depend upon each other.

d) A male student is selected find the probability that he studies Economics

$P(\frac { E }{ M } )\quad =\quad \frac { P(E\quad \cap \quad M) }{ P(M) } \quad =\quad \frac { \frac { 41 }{ 144 } }{ \frac { 91 }{ 144 } } \quad =\quad \frac { 41 }{ 91 }$