# Probability - Aptitude Tricks, Formulas & Shortcuts

This is an important topic for all of those students appearing in competitive exams like Placement test, Bank PO, CAT, XAT, SNAP etc. because the questions from probability will test your counting skills such as permutation and combination too. Once you are really skilled in combinatorics (Basic counting methods, permutation and combination) you can easily solve the aptitude questions from this area by using common sense. In this article, we provide - basic terminologies of probability, important formulas, basic and advanced concepts, Baye's Theorem, probability aptitude question solving tricks.

The chance to occur an event is a simple and basic definition for the term probability.
So, probability of an event = No of favourite outcomes of the event / Total number of possible outcomes

## Probability : Basic Terminologies, Formulas & Concepts

1. Deterministic and probabilistic (or Random experiments)
If the outcome of an operation is certain, then the operation is called a deterministic experiment, then there is no scope of probability.
E.g. Direction of raising the sun is a deterministic natural phenomenon. Most of the Laboratory experiment are deterministic like simple pendulum, glass prism etc.
If the outcome of any operation is not predictable, then the operation is not deterministic, then it is called a probabilistic (or random) experiment.

E.g. winning of a team in a cricket match, Getting rain in a particular place etc.

2. Random Experiment:

If an experiment is not deterministic and it gives any of the possible outcomes, then the experiment is called a random experiment.

E.g. tossing of a fair coin, Rolling of an unbiased die.

3. Sample Space:

The set of all possible outcomes of a random experiment is called sample space.

E.g. When rolling n unbiased die, then there are 6 possible outcomes as the result.The set of all possible outcomes can be arranged in a set in the form S={1,2,3,4,5,6}

This set is the sample space of the event of rolling a die.

4. Event

Any subset of a sample space is called an event.

e.g. In a flipping of coin ,the event of getting a head can be considered as {H}.This event is a subset of the sample space{H,T} of the experiment.

5. Elementary Event

Basically it is a singleton subset of the sample space of any operation. An event containing only a particular outcome of an operation is called an elementary event.

E.g. {1},{2},{3},{4},{5},{6} are the six different elementary events of the operation "rolling an unbiased die".

6. Compound Events

If an event contains more than one element from the sample space of an operation is called a compound event.

E.g. an event of getting even numbers in the rolling of a die is {2, 4, 6} and this is a compound event.

7. Impossible Event

Basically it is an empty event or it is the not happening condition (i.e. it is an event which can never occur under any circumstances).

E.g. Event of getting a two digit number in the rolling of die is {} or

8. Sure Event

It is an event which is sure to occur. It is representing the sample space itself.

E.g. the event of getting a single digit number in throwing a die is a sure event.

9. Equally Likely Event

If there is not any preference for any of the possible outcomes in an event (experiment) or the chance of occurrence of the outcome is equally distributed in all the possible outcomes in an event is called an equally likely event.

E.g. while flopping a fair coin, the chance to get a head or tail is equally distributed.

10. Complementary Events

An event is representing the complementary of any other event E is called the complementary event and it is denoted by Ec

E U Ec = sample space

E.g. Let E be the event of getting an odd number in through a die and the Ec is the event of not getting an odd number in same operation.

11. Mutually Exclusive Events

If two events E1 and E2 are totally different, then E1 and E2 are called mutually exclusive events. Or in another words, one of the events always prevents the happening of the other.

E1 ∩ E2 = ψ

E.g. consider the operation of flipping a coin .Let E1 be the event of getting a head and E2 be the event of getting a tail.

E1= {H}
E2= {T}
E1 ∩ E2 = ψ

Consider another example of drawing a card from a well shuffled pack of playing cards.

E1 - An event of getting a heart.

E2 - An event of getting a spade.

Here E1c is the complementary of E1 but E2 is not a complementary of E1 because

E1 ∪ Ec = sample space, but E1 ∪ E2 ≠ sample space

E1 and E2 are mutually exclusive events.

i.e. E1 ∩ E2 = ∅

12. Probability of an event

Let E be an event and Ec is its complementary event. If S is the sample space, then the probability of an event E,

p(E) = n(E) / n(S)

p(Ec) = n(Ec) / n(S)

= n(S) - n(E) / n(S)

= 1 - (n(S) / n(E))

= 1 - p(E) ie. p(Ec) = 1 - p(E)

or

p(E) + p(Ec) = 1

Probability of an event E normally represented by 0, proper fraction or 1.

i.e. 0 ≤ p(E) ≤ 1

If p (E) = 0 then the event is an impossible event.

If p (E) = 1 then the event is a certain event.

In general, for any event E;

0 ≤ n(E) ≤n(S)

0 ≤ n(E) / n(S) ≤ 1

0 ≤ p(E) ≤ 1

13. Odds in favor of an event

It is basically a ratio of the probability of an event E and the complementary event Ec.

i.e. odds in favor of an event E = p(E) / p(Ec)

E.g. Odds in favor of getting prime number in rolling of a die is

n({2,3,5}) / n({1,4,6}) = 3 / 3 = 1 V 1 : 1

14. Odds against an event

It is the ratio of the probabilities of the complementary of an event to the event itself.

i.e. odds against an event E is p(Ec) / p(E)

E.g. Let E be the event of getting a multiple of 3 in rolling a die.

Then E = {3,6} and Ec = {1,2,4,5}

Odds against the event,

E = p(Ec) / p(E)

= n(Ec) / n(E)

= 4 / 2

= 2:1

## Probability: Operations

### Addition theorem for two events

If a and B are two events associated with the same random experiment (i.e. both the events from the same sample space S) then

P (A ∪ B) = p(A) + p(B) - p(A ∩ B)
i.e. p(A or B) = p(A) + p(B) - p(A and B)
If A and B are two mutually exclusive events, i.e. A ∩ B = ∅ , then p (A ∩ B) = 0
P(A or B) = p(A) + p(B)
E.g. When drawing a card from a well shuffled pack of playing cards, what is the probability of getting a jack or a spade?
Solution:
Here we need to find p(jack or a spade)
= 4 / 52 + 13 / 52 - 1 / 52

= 16 / 52

= 4 / 13

### Addition theorem for three events

If A, B and C are three events associated with the same random experiment, then
p(A ∪ B ∪ C)= [p(A) + p(B) + p(C)] - [p(A ∩ B) + p(B ∩ C) + p(A ∩ C)] + p(A ∩ B)
If A, B and C are three mutually exclusive events, then p(A ∪ B ∪ C)= p(A) + p(B) + p(C)
E.g. while selecting a number from the set of first hundred natural numbers. What is the probability of getting a number which is a multiple of 3, 4 or 5?
p(multiple of 3, 4 or 5) = [ p(multiple of 3) + p(multiple of 4) + p(multiple of 5) ] - [ p(multiple of 3 and 4) + p(multiple of 4 and 5) + p(multiple of 3 and 5) ] + p(multiple of 3,4 and 5)

Number of multiples of 3 = 33
Number of multiples of 4 = 25
Number of multiples of 5 = 20
Total = 78

Number of, Multiple of 3 and 4 = multiple of 12 = 8
Multiple of 4 and 5 = multiple of 20 = 5
Multiple of 3 and 5 = multiple of 15 = 6
Total = 19

Number of multiples of 3, 4 and 5 = multiples of 60 = 1
p = 78 / 100 - 19 / 100 + 1 / 100
= 60 / 100
= 3 / 5

### Probability of difference of two events

If A and B are two events associated with the same random experiment, then

p(A ∪ B ∪ C)= [ p(A) + p(B) + p(C) ] - [ p(A ∩ B) + p(B ∩ C) + p(A∩ C) ] + p(AB)
p(A - B) = p(A) - p(A ∩ B)
p(B - A) = p(B) - p(A ∩ B)
p(A - B) = p(A) - p(A ∩ B)
p(A - B) = p(B) - p(A ∩ B)

### Probability of exactly one out of two events occurs

If A and B are two events associated with the same random experiment.

p(A-B) = p(A) - p(A ∩ B)
p(B-A) = p(B) - p(A ∩ B)
p[(A-B) or (B-A)] = p(A) + p(B) - 2p(A ∩ B)
Or p(A ∪ B) - p(A ∩ B)

### Probability of exactly one of the three events occurs

If A, B and C are three events associated with the same random experiment then

p[exactly one of A,B and C] = [ p(A) + p(B) + p(C) ] - 2[p(A ∩ B) + p(B ∩ C) + p(A ∩ C)] - 3p(A ∩ B ∩ C)

• p[exactly two of A,B and C occur] = p(A ∩ B) + p(B ∩ C) + p(A ∩ C) - 3p(A ∩ B ∩ C)
• p[at least two of A,B and C occur] = p(A ∩ B) + p(B ∩ C) + p(A ∩ C) - 2p(A ∩ B ∩ C)

Now we will look into the advanced application of probability concepts. Sometimes it is quiet difficult to understand whether the question required an advanced kind of application such as 'Conditional Probability', so an aspirant should aware about the exact situations in which these advanced results are applicable.

The advanced probability concepts we are going to look into are

• Conditional probability
• Independent events
• Law of total probability
• Baye's theorem

### Conditional Probability

Let A and B are any two events associated with a random experiment. The probability of the occurrence of event A when the event b has already occurred is called the "conditional probability of the occurrence of A when B is given" and is denoted as P(A/B) read as "Probability of A given B".

The conditional probability P(A/B) is meaningful only when P(B) ≠ 0, i.e. when is not an impossible events.

Eg: Let us consider the experiment of flipping two coins simultaneously.

S = {(H, H), (H, T), (T, H), (T, T)}
Let, A = {(H, H), (H, T)}
B = {(H, H), (T, H)}
C = {(H, H), (H, T), (T, H)}
P(A) = 2/4 = 1/2
P(B) = 1/2
P(C) = 3/4

A/B is the occurrence of the event A with condition that the event B already occurred.

P(A/B) = n{(H,H)} / n{(H,H), (T,H)} = 1/2
i.e. P(A/B) = P(A ∩ B) / P(B)

Similarly, P(B/C) = 2/3

and P(A/C) = 2/3

P(A/B) = P(A ∩ B)/P(B) = P(B/A) = P(A ∩ B) / P(A)
P(A ∩ B) = P(A).P(A/B) or P(B). P(A/B)

### Probability - Independent Event

If the occurrence of one event doesn't depend upon the occurrence of another event, then both the events are called independent events.

Consider an urn, which contains 5 red balls and 4 green balls.

Let A1 - an event of selecting one red ball in first draw.

Let A2 - an event of selecting one green ball in second draw.

If the selection of balls one after another with replacement, then the occurrence of the event A1 can't after the occurrence of the event A2.

There for, A1 and A2 are two independent events.

If the selection of balls one after another without replacement, then the occurrence of A1 will after the occurrence of A2.

A1 and A2 are not two independent events.

There for, Two events A and B associated with a random experiment are said to be independent if,

P(A ∩ B) = p(A).p(B)

There for, p(A/B) = p(A) and p(B/A) = p(B)

In general, if A1, A2, A3 , ....., An are independent events associated with a random experiment then

p(A1 ∩ A2 ∩ A3 ..... ∩ An) = p(A1).p(A2)......p(An)

If A1, A2, A3 ..... An are 'n' independent events associated with the same random experiment, then

p(A1 ∪ A2 ∪ A3 ...... ∪ An) = 1- [p(A1).p(A2) .... p(An)]

### Probability : Baye's Theorem

If E1,E2 ..... En are mutually exclusive and exhaustive with the same random experiment with non zero probabilities,

i.e. p(Ei) ≠ 0, for i 1,2,...n)

then for any event A which is a subset of total sample space [S= E1 ∪ E2 ... ∪ En] such that p(A) > 0, then

p(Ei/A) = P(Ei).P(A/Ei) / Σj = 1 to n P(Ej).P(A/Ej)

Example:

There are two bags contains a certain number of gold coins and silver coins. 3 gold coins and 5 silver coins in Bag1, 6 gold coins and 2 silver coins in Bag2.A coin is drawn from one of the two bags, the coin drawn is gold. What is the probability that it has drawn from Bag2?

Solution:

Let A and B are the events of selecting Bag1 and Bag2 respectively.

p(A) = 1/2

p(B) = 1/2

Let C be the event of selecting a gold coin.

Then, p(C/A) = p(selecting a gold coin from Bag1) = 3/8

Similarly p(C/B) = 6/8

Coin selected is gold, then the probability of the selection done from Bag2 is p(B/C)

i.e. p(B/C) = p(B).p(C/B) / (p(A).p(C/A) + p(B).p(C/B))

### Law of total probability

Before discussing Bayes theorem we have to discuss law of total probability.

E1,E2 ..... En are n mutually exclusive and exhaustive events associated with a random experiment and an arbitrary event a is occurs with E1 or E2 or ... or En then

p(A) = p(E1).p(A/E1) + p(E2).p(A/E2) + ... + p(En).p(A/En)

Example:

There are 5 boys and 6 girls in class A and 3 boys and 4 girls in class B. Two students are shifted from class A to class B. Then selecting two students from class B, what is the probability that selecting one boy and one girl?

Solution:

There are three probabilities for the shifting of two students from class A to class B

• shifting two boys
• shifting two girls
• shifting one boy and one girl

Let E1 be the event of shifting two boys from class A to class B.

E2 be the event of shifting two girls from class A to class B.

E3 be the event of shifting one boy and one girl from class A to class B.

N be the event of shifting one boy and one girl from class A to class B.

p(E1) = 5C2 / 11C2 = 2/11

p(E2) = 6C2 / 11C2 = 3/11

p(E3) = 5C1 * 6C1 / 11C2 = 6/11

If the event E1 has already occurred. i.e. two boys are shifted from class A to class B. Then the number of boys and girls in class B are 5 and 4 respectively.

p(N/E1) = probability of selecting one boy and one girl from class B after the occurrence of event E1.

= 5C1 * 4C1 / 9C2 = 5/9

Similarly,

p(N/E2) = 3C1 * 6C1 / 9C2 = 1/2

p(N/E3) = 4C1 * 5C1 / 9C2 = 5/2

The total probability

p(N) = p(E1).p(N/E1) + p(E2).p(N/E2) + p(E3).p(N/E3)

= [2/11 * 5/9] + [3/11 * 1/2] + [6/11 * 5/9]

= 10/99 + 3/22 + 30/99

= 40 / 99 + 3 / 22

= (80 + 27) / 198

= (107) / 198

### Probability : Expected value

In a random experiment, if each outcome is associated with a value x, then the expected value of the experiment is

Σ PiXi / Σ Pi

There for, Σ Pi is the sum of probabilities of the distinct events associated with the given random experiment.

Σ Pi = 1
Expected value = Σ PiXi

Example

A game coin throwing is based on an arrangement of 9 boxes arranged in a 3 * 3 square of different boxes form as in the given diagram. Each box addressed with different colours. There are 4 boxes in red color, 3 boxes in green color and remaining two boxes are in yellow color.

 R G R Y R G G Y R

When the participant is throwing a coin, if the coin is fall into any of the green box, the participant will get Rs.10.If the coin is fell into an yellow box he won't get or loss any amount. If the coin fell into a red box, the participants will loss Rs.7. What is the expected value of the game?

Solution:

Let 'R' be the event of the coin fell into a red box.

'G' be the event of the coin fell into a green box.

and 'Y' be the event of the coin fell into an yellow box.

P(R) = 4/9

P(G) = 3/9

P(Y) = 2/9

Expected value = (P(G) * 10) + (P(Y) * 0) + (P(R) * (-7))

= 3/9 * 10 - 4/9 * 7

= 30/9 - 28/9

= 2/9 tends to 0.22