Independent and Mutually Exclusive Events

Consider the task of tossing a die and flipping a coin at the same time. Let Hn denote that the coin landed on heads and n was tossed on the die and let Tn denote that the coin landed on tails and n was tossed on the die.

The sample space for this experiment is

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Let A be the event that an even number was tossed on the die.
Let B be the event that a number divisible by 5 was tossed on the die.
Let C be the event that the coin landed on heads.

Let's represent these events as sets.

A = {H2, H4, H6, T2, T4, T6}
B = {H5, T5}
C = {H1, H2, H3, H4, H5, H6}

P(A) = 6/12 = 1/2
P(B) = 2/12 = 1/6
P(C) = 6/12 = 1/2

Since there is no number between 1 and 6 that is both even and divisible by 5 the events, A and B, cannot take place simultaneously. A and B are said to be mutually exclusive events or in other words their sets do not intersect.

The probability that A or B takes place may be calculated as follows:
P(A or B) = P(A) + P(B) = 1/2 + 1/6 = 4/6

Since it is possible for an even number to be tossed on the die and for the coin to land on heads simultaneously then A and C are not mutually exclusive. In fact A ∩ C is a non-empty set,  A ∩ C = {H2, H4, H6}.

Two events are considered independent when the outcome of one does not influence the outcome of the other. In other words, the success of one event does not affect the probability of the other event. A and C are independent events and B and C are independent events since the side the coin lands on is in no way affected by the number tossed on the die.

Since A and C are independent, the probability that A and C take place simultaneously may be calculated as follows:
P(A and C) = P(A) × P(C) = 1/2 + 1/2 = 1/4

How do we calculate the probability of A or C? Since A and C are independent we use the following formula:
P(A or C) = P(A) + P(C) - P(A and C) = 1/2 + 1/2 - 1/4 = 3/4


Test yourself, are events B and C mutually exclusive? How would you calculate P(B or C)?

Proof by Contradiction

One strategy for proving the validity of an argument is Proof by Contradiction which is based on the Reductio ad Absurdum (Latin for "reduction to the absurd") argument.

In this type of proof, we usually assume the negation of the conclusion. We use this assumption together with the other premises to derive a result. This result is either already known to be false or can be shown to be false. Since the result cannot both be true and false then it follows that the initial assumption must be false i.e. the conclusion holds.

Let's use propositional logic and the Proof by Contradiction approach to prove the following argument.

Either Veronica is not from Riverdale or Dilton is not smart.
Dilton is smart.
If the Lodges did not move then Veronica is from Riverdale.
Therefore the Lodges moved.

Let R be the proposition "Veronica is from Riverdale."
Let S be the proposition "Dilton is smart."
Let M be the proposition "The Lodges moved."

Distractions

According to Google Dictionary:

dis·trac·tion

noun /disˈtrakSHən/ distractions, plural A thing that prevents someone from giving full attention to something else - the company found passenger travel a distraction from the main business of moving freight A diversion or recreation - there are plenty of distractions such as sailing

Mobile games, text messages, IM, email, the WWW, friends - all possible distractions to a student  (definition 1) while in class. The further you sit to the back of a classroom, you think, the less noticeable you are and the more tempted you are to indulge in these distractions. Some teachers actually pay more attention to those in the back rows for this very reason.

Why do you choose to be distracted when you've already done the hard part? You've made it to class. You're here. At the very least you could listen. And while you're at it, why not take some notes and participate? It's all been (or being) paid for - the tuition, the transportation, the books, the stationery. Somebody's paying for them, if not you, then your parents or the institution or the government or some sponsor. So much can be gained during these contact hours with your teacher and your peers. Listen. Engage. Learn.

The truth is, mobile games, text messages, IM, email, the WWW, friends are distractions (definition 2) that will still be available after class for you to enjoy.

Proofs and Equality of Equivalence Classes

Courses in Number Theory require students to prove various theorems. In some cases, as you construct your proof you will find that you need to show that two equivalence classes are the same. In these cases it is useful to remember the relation that is involved. If two elements are related then these elements belong to the same equivalence class.

Consider the following proof exercise:

Show that \(x+0=0\) for every \(x \epsilon  \mathbb{Z}\)

Proof:

Let \(x = [a,b]\) and recall that \(0 = [1,1]\).

Recall the definition of integer addition i.e. \([r,s] + [t,u] = [r+t, s+u]\).

So,  \(x + 0 = [a, b] + [1, 1] = [a+1, b+1]\).

We need to show that \([a+1, b+1]  = [a,b]\).

Recall that  \(\mathbb{Z}\) is the set of equivalence classes of R where R is the relation on \(\mathbb{X}\) such that \((a,b)R(c,d)\) means \(a+d=b+c\) and \(\mathbb{X}\) is the set of ordered pairs \((a,b)\) of natural numbers.

\((a+1,b+1)R(a,b)\) since \(a+1+b = b+1+a\)

Thus, \([a+1,b+1]=[a,b]\) and the result holds.