I received quite a few requests for putting up a write up on probability. Today I finally decided to pen it down.

Probability tells how likely something is to happen. It tells the chance that an event will occur.

If I have to toss a coin, I cannot be certain that a head or a tail will occur. I can only talk about what is the chance for a head or a tail to occur.

Assuming that the coin will fall on either side only and not stand vertical ;), there are only two possible outcomes when I toss a coin – head or tail. Therefore, both are equally likely (unless of course the coin is biased). probability of getting a head is ½ and probability of getting a tail is also ½. In other words there is a 50-50 chance for both the outcomes.

When a single die (Yes, die. Actually dice is the plural form of die) is rolled, there are 6 possible outcomes – 1, 2, 3, 4, 5 & 6 and all the six outcomes stand an equal chance of occurring. So each outcome has a probability of ⅙.

In general, probability of an event = no. of favourable outcomes/total no. of possible outcomes

**0≤P(E)≤1**

It might have occurred to you by now that probability of any event will always lie between 0 and 1. 0 being impossible and 1 being absolutely certain.

For example, probability of getting a 9 on throwing a die is 0 because it is impossible (at least in the case of a normal die) and probability of getting a number less than 10 is 1 because it is certain.

### Determining the total number of outcomes

If we have to toss 1 coin there are only two outcomes.

If we toss 2 coins together, there are 4 outcomes – HH TT HT TH

If we toss 3 coins together, then?

HHH TTT HHT HTH HTT TTH THT THH – 8 outcomes.

So how do we compute this?

Every time you increase a coin, the total number of outcomes doubles.

What if it is a die? Rolling 1 die results in 6 outcomes; 2 die in 36 outcomes; 3 die in 216 outcomes. Phew!!

Can we generalize this? Yes we can.

**Total number of outcomes** = (Number of outcomes per object) to the power ^{Number of objects}

That means if i am tossing 8 coins together, total no. of outcomes that I will get is 2 to the power 8 (2^8)

Similarly, total no. of outcomes on rolling 10 die will be 6 to the power 10 (6^10).^{
}

### Odds in Favour and Odds Against

You will find ample usage of these two terms in probability related questions. Odds in probability defines a ratio between favourable and unfavourable events.

Odds in favour = No. of favourable cases / No. of unfavourable cases

Odds against = No. of unfavourable cases / No. of favourable cases

Example: Two fair coins are tossed. Find the odds in favour of getting at least one head.

Solution: No. of favourable cases= 3 (HT TH HH)

No. of unfavourable cases= 1 (TT)

Hence, Odds in favour of getting at least one head ( i.e. one or more) = ¾.

### Mutually exclusive and exhaustive system of events

If two events are said to be mutually exclusive then if one happens, the other cannot happen and vice-versa. Basically, the events cannot occur simultaneously.

In general **P(A or B) = P(A) + P(B) – P(AB)**

If two events are mutually exclusive then P(AB)=0

For example, in rolling a die

A: the event that the number is odd

B: the event that the number is even

C: the event that the number is a multiple of 3

Let us see what each of these events encompass.

A: {1 3 5} B: {2 4 6} C: {3 6}

Events A and B are mutually exclusive because a number can be either odd or even and not both simultaneously. Basically, there is no common element between A and B.

Events A and C are not mutually exclusive or disjoint since they have a common outcome – 3.

Events A{1,3,5} and B{2,4,6} together also form an **exhaustive system of events** as together they have included all possible outcomes and there is no outcome beyond A and B for the event ‘rolling a die’.

### Additional Law of Probability

In case of mutually exclusive events, probability that either event A or event B will occur in a single trial is given by: **P(A or B) = P(A) + P(B)**. This can also be compared with set theory.

Similarly, **P(neither A nor B) = 1- P(A or B)**.

In case of 3 events,

P(A or B or C) = P(A) + P(B) + P(C) – P(A and B) – P(B and C) – P(A and C) + P(A and B and C)

If A, B and C are mutually exclusive then P(A and B) = P(B and C) = P(A and C) = P(A and B and C) = 0

### Independent Events

Two events are independent if the occurrence of one has no effect on the occurrence of the other.

For example, on rolling a die and tossing a coin together:

A: the event that the number 5 turns up

B: the event that tail turns up

Here both A and B are completely independent of each other. Getting a 5 cannot have any impact on getting a tail.

Take another example.

Suppose a bag contains x red balls and y black balls and two balls are drawn out.

There can be two cases here –

one, the first ball is replaced and then the second ball is drawn;

two, the first ball is not replaced and the second ball is drawn out of the remaining balls.

If the ball drawn in the first draw is not replaced back in the bag, then two events of drawing the ball are dependent because first draw of the ball would determine the probability of drawing the second ball. If the ball drawn in the first draw is replaced back in the bag, then two events are independent because first draw of a ball will have no effect on the second draw.

### Multiplication law of probability

If the events A and B are independent, then P(A and B) = P(A) x P(B).

### Conditional Probability

The concept of conditional probability is used only in case of dependent events.

Let A and B be two dependent events, then probability of occurrence of event A when B has already occurred is given by P(A|B) = P(AB)P(B).

Let us understand this with an example. A pair of dice is thrown simultaneously. We have to find the probability that the sum obtained is 8 when there is an even number on the 1st die.

A: Sum of 8 is achieved.

A= {(2,6), (3,5), (4,4), (5,3), (6,2)}

B: we get an even number on the 1st die.

B= {(2,1), (2,2), (2,3),…,,(4,1), (4,2), (4,3),…, (6,1), (6,2), (6,3),…(6,6)}

Total no. of outcomes= 62=36

P(A)= 5/36 and P(B)=18/36

P(A|B) = P(AB)P(B)= (2,6) (4,4) (6,2)P(B)= 3/3618/36=3/18.

I have covered all the basics that you need to know to understand thoroughly the basics of probability.

This article is written from the viewpoint of competitive exam takers. Hence I have put in what they need to know. I have avoided detailed definitions of every single term to maintain crispness. Also I have deliberately avoided explanations using too much of set theory and series and functions notations to keep it simple and easy to digest.

Hope you like it. Waiting for your feedback!

No. of favourable cases= 3 (HT TH HH)

No. of unfavourable cases= 1 (TT)

Hence, Odds in favour of getting at least one head ( i.e. one or more) = ¾.

but

Odds in favour = No. of favourable cases / No. of unfavourable cases

then why not =3/1

you said,

total no of outcomes of 8 coins= 28

and of 10 dice= 610

I didn’t get this…plz explain

Hi Nilesh

Thanks for writing in. It is a small mistake of ignoring the superscript while posting. It will be 2 to the power 8 and 6 to the power 10. correction made. check it out.

good work on basics.

Murali – Thanks for appreciating. We’ll come with more such articles in near future. Here’s one already posted on Permutation & Combibation – http://www.quickermaths.com/permutations-and-combinations-part-1-fundamental-principle-of-counting/