When MS and MBA applicants ask us – ‘*What are my chances of getting into Harvard?*‘ or ‘*What’s my probability of getting scholarships from Oxford?*‘ we get tongue-tied. There are so many variables at play, it’s difficult to give an accurate answer.

But when you get probability questions in your GRE and GMAT exam syllabus, you don’t have to get flummoxed. Understanding the basic rules and formulas of probability will help you score high in the entrance exams.

## Meaning and definition of Probability

As the Oxford dictionary states it, Probability means ‘The extent to which something is probable; the likelihood of something happening or being the case’.

In mathematics too, probability indicates the same – the likelihood of the occurrence of an event.

Examples of events can be :

- Tossing a coin with the head up
- Drawing a red pen from a pack of different coloured pens
- Drawing a card from a deck of 52 cards etc.

Either an event will occur for sure, or not occur at all. Or there are possibilities to different degrees the event may occur.

An event that occurs for sure is called a Certain event and its probability is 1.

An event that doesn’t occur at all is called an impossible event and its probability is 0.

This means that all other possibilities of an event occurrence lie between 0 and 1.

This is depicted as follows:

**0 <= P(A) <= 1**

where A is an event and P(A) is the probability of the occurrence of the event.

This also means that a probability value can never be negative.

Every event will have a set of possible outcomes. It is called the ‘sample space’.

Consider the example of tossing a coin.

When a coin is tossed, the possible outcomes are Head and Tail. So, the sample space is represented as {H, T}.

Similarly when two coins are tossed, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

The probability of head each time you toss the coin is 1/2. So is the probability of tail.

#### Basic formula of probability

As you might know from the list of GMAT maths formulas, the Probability of the occurrence of an event A is defined as:

**P(A) = (No. of ways A can occur)/(Total no. of possible outcomes)**

Another example is the rolling of dice. When a single die is rolled, the sample space is {1,2,3,4,5,6}.

What is the probability of rolling a 5 when a die is rolled?

No. of ways it can occur = 1

Total no. of possible outcomes = 6

So the probability of rolling a particular number when a die is rolled = 1/6.

## Compound probability

Compound probability is when the problem statement asks for the likelihood of the occurrence of more than one outcome.

#### Formula for compound probability

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

where A and B are any two events.

P(A or B) is the probability of the occurrence of atleast one of the events.

P(A and B) is the probability of the occurrence of both A and B at the same time.

### Mutually exclusive events:

Mutually exclusive events are those where the occurrence of one indicates the non-occurrence of the other

OR

When two events cannot occur at the same time, they are considered mutually exclusive.

__Note:__ For a mutually exclusive event, P(A and B) = 0.

**Example 1: **What is the probability of getting a 2 __or__ a 5 when a die is rolled?

__Solution:__

Taking the individual probabilities of each number, getting a 2 is 1/6 and so is getting a 5.

Applying the formula of compound probability,

Probability of getting a 2 **or **a 5,

P(2 or 5) = P(2) + P(5) – P(2 and 5)

==> 1/6 + 1/6 – 0

==> 2/6 = 1/3.

**Example 2: **Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

__Solution:__

We need to find out P(B or 6)

Probability of selecting a black card = 26/52

Probability of selecting a 6 = 4/52

Probability of selecting both a black card and a 6 = 2/52

P(B or 6) = P(B) + P(6) – P(B and 6)

= 26/52 + 4/52 – 2/52

= 28/52

= 7/13.

### Independent and Dependent Events

#### Independent Event

When multiple events occur, if the outcome of one event __DOES NOT__ affect the outcome of the other events, they are called independent events.

Say, a die is rolled twice. The outcome of the first roll doesn’t affect the second outcome. These two are independent events.

**Example 1: **Say, a coin is tossed twice. What is the probability of getting two consecutive tails ?

Probability of getting a tail in one toss = 1/2

The coin is tossed twice. So 1/2 * 1/2 = 1/4 is the answer.

Here’s the verification of the above answer with the help of sample space.

When a coin is tossed twice, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

Our desired event is (T,T) whose occurrence is only once out of four possible outcomes and hence, our answer is 1/4.

**Example 2: **Consider another example where a pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 blue pens and 1 black pen?

**Solution**

Here, total number of pens = 9

Probability of drawing 1 blue pen = 4/9

Probability of drawing another blue pen = 4/9

Probability of drawing 1 black pen = 3/9

Probability of drawing 2 blue pens and 1 black pen = 4/9 * 4/9 * 3/9 = 48/729 = 16/243

#### Dependent Events

When two events occur, if the outcome of one event affects the outcome of the other, they are called dependent events.

Consider the aforementioned example of drawing a pen from a pack, with a slight difference.

**Example 1:** A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from the pack, __NOT__ replaced and then another pen is drawn. What is the probability of drawing 2 blue pens and 1 black pen?

**Solution:**

Probability of drawing 1 blue pen = 4/9

Probability of drawing another blue pen = 3/8

Probability of drawing 1 black pen = 3/7

Probability of drawing 2 blue pens and 1 black pen = 4/9 * 3/8 * 3/7 = 1/14

Let’s consider another example:

**Example 2:** What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, __without__ replacement.

Probability of drawing a king = 4/52 = 1/13

After drawing one card, the number of cards are 51.

Probability of drawing a queen = 4/51.

Now, the probability of drawing a king and queen consecutively is 1/13 * 4/51 = 4/663

## Conditional probability

Conditional probability is calculating the probability of an event given that another event has already occured .

The formula for conditional probability P(A|B), read as P(A given B) is

**P(A|B) = P (A and B) / P(B)**

Consider the following example:

**Example:** In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?

**Solution**

P(M and S) = 0.40

P(M) = 0.60

P(S|M) = P(M and S)/P(S) = 0.40/0.60 = 2/3 = 0.67

### Complement of an event

A complement of an event A can be stated as that which does NOT contain the occurrence of A.

A complement of an event is denoted as P(A^{c}) or P(A’).

**P(A**

^{c}) = 1 – P(A)**or it can be stated, P(A)+P(A ^{c}) = 1**

For example,

if A is the event of getting a head in coin toss, A^{c} is not getting a head i.e., getting a tail.

if A is the event of getting an even number in a die roll, A^{c }is the event of NOT getting an even number i.e., getting an odd number.

if A is the event of randomly choosing a number in the range of -3 to 3, A^{c} is the event of choosing every number that is NOT negative i.e., 0,1,2 & 3 (0 is neither positive or negative).

Consider the following example:

**Example:** A single coin is tossed 5 times. What is the probability of getting at least one head?

__Solution:__

Consider solving this using complement.

Probability of getting no head = P(all tails) = 1/32

P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.

## Sample Probability questions with solutions

#### Probability Example 1

What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled.

**Solution**

Let the event of the occurrence of a number that is odd be ‘A’ and the event of the occurrence of a number that is less than 5 be ‘B’. We need to find P(A or B).

P(A) = 3/6 (odd numbers = 1,3 and 5)

P(B) = 4/6 (numbers less than 5 = 1,2,3 and 4)

P(A and B) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)

Now, P(A or B) = P(A) + P(B) – P(A or B)

= 3/6 + 4/6 – 2/6

P(A or B) = 5/6.

#### Probability Example 2

A box contains 4 chocobars and 4 ice creams. Tom eats 3 of them one after another. What is the probability of sequentially choosing 2 chocobars and 1 icecream?

**Solution**

Probability of choosing 1 chocobar = 4/8 = 1/2

After taking out 1 chocobar, the total number is 7.

Probability of choosing 2nd chocobar = 3/7

Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3

So the final probability of choosing 2 chocobars and 1 icecream = 1/2 * 3/7 * 2/3 = 1/7

#### Probability Example 3

When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8.

**Solution**

Let the event of getting a greater number on the first die be G.

There are 5 ways to get a sum of 8 when two dice are rolled = {(2,6),(3,5),(4,4), (5,3),(6,2)}.

And there are two ways where the number on the first die is greater than the one on the second given that the sum should equal 8, G = {(5,3), (6,2)}.

Therefore, P(Sum equals 8) = 5/36 and P(G) = 2/36.

Now, P(G|sum equals 8) = P(G and sum equals 8)/P(sum equals 8)

= (2/36)/(5/36)

= 2/5

## Probability Quiz: Sample probability questions for practice

**Problem 1:** Click here

**Answer 1:** Click here

**Problem 2:** Click here

**Answer 2:** Click here

Learn how to solve:

– Simple and compound interest problems

– Speed, distance and time problems

– Ratio and proportion

– List of Maths Formulas

**Become a Probability & Statistics Master**

a number of people gave a hat check girl one hat. suppose all the tickets got misplaced, so all the hat were given back randomly.

a) if its 2 people determine the probability at least one person got their hat returned.

b) if its 3 people determine the probability at least one person got their hat returned.

c) if its 4 people determine the probability at least one person got their hat returned.

d) if its 5 people determine the probability at least one person got their hat returned.

Hi I’m Algia and I need help in solving this problem, can you help me please.

a) 1-(0.5)=0.5

b) 1-(0.667*0.5)=0.667

c) 1-(0.75*0.667*0.5)=0.75

d) 1-(0.8*0.75*0.667*0.5)=0.8

Really nice sequence.

P(h’)=1-P(h) etc. At least one hat is correctly returned is compliment that no hat is returned correctly.

(n-1)/n

The math here is totally wrong

Feel free to enlighten!

you make the assumption the events are independent, they are not, if you there are only 2 people, it is impossible for just 1 person to be given the right hat, it must be both people.

You make the assumption of independence, which does not hold in this question. instead we have to use the inclusion-exclusion principle.

For a) the probability that at least 1 person gets the got their hat returned. A is the event the 1st person gets their hat and B is the event the 2nd person gets their hat. We want to find P(A U B) = ?

P(A U B) = P(A) + P(B) – P(A n B)

P(A) = 1/2

P(B) = 1/2

P(A n B) = 1/4

P(A U B) = 1/2 + 1/2 – 1/4 = 3/4

Think it in this way. Representation here: Got Hat – G, Didn’t Get Hat – D. If there are 2 people, the total possible outcomes would be = 2^2 = 4 with them being {(G,G),(D,D),(G,D),(D,G)}. Now, probability that at least one person got hat returned = 1 – P(no one gets hat) = 1 – 1/4 = 3/4.

Similarly, in 3 people case, probability that at least one person gets hat = 1 – P(no one gets hat) = 1 – 1/8 = 7/8.

and so on…

General Formula derived for ‘n’ people case : P(At least one person gets hat returned) = 1 – P(no one gets hat) = (1 – 1/2^n)

Hi

Last question must be 212/216 right ?

Tell me the way u did that sol. Plzz…

yes its must be 212/216

I think it should be 212/216

Bcz we have 4 number of event to getting a number of sum less than 5

{(1,1,1),(1,1,2),(1,2,1),(2,1,1)}

It means p(e) = 4/216

Nd getting a number of sum at least 5

Is

1-4/216=212/216

you are telling a wrong way so stop telling me your bloody wrong answer it’s a correct answer it’s should not be 212/216..

@Emmanuel, I don’t think you can isolate the two variables like that since, as you mentioned, these are dependent events. The total number of ways of returning the hat is n! and the total number of ways of returning the hats such that no one gets their own one is (n-1)!. Therefore, the probability that at least one person gets their own hat is P(X>0) = 1-P(X=0) = 1 – (n-1)!/n! = 1 – 1/n.

A woman bought 5basket of tomatos each costing 1250naira,in her discovery she observe that 90% of the tomatos where damage resulting to a loss of 510naira.(a)what is the probability of obtaining an average of 50 if the cost per bag is 50 above the cost?(b)what will be the actual price for selling the tomato at cost plus(+) 25%?

Simple way:

Q: A die is rolled thrice. What is the probability that the sum of the roll is at least 5?

A: At least 5 is equal or greater 5 ===> which is P(x> or =5)= 1-P(x<5)

Then: P(x<5)= 4/216. . .taken thrice rolls =6*6*6=216

Therefore: 1-P(x 53/54

This is incorrect. Correct answer is:

P(x<5) = 3

Therefore Answer = 1 – 3/216 = 213/216.

the personal director of a company wishes to select applicant for advanced training without regard to sex. let “W” denotes women and “M” denotes men and the pattern of arrival be M WWW MMM WW M WWW MMMM W M W MM WWW MM W MMMM WW M WW MMMM WW M WWWW MM WW M W WW. will you conclude that the applicants have arrived in a random fashion?

pavement.before any 250 m length of a pavement is accepted by the state highway department,the thickness of a30 m s mointored by an altrasonic to verify compliance to specification .each section is rejected if a measurment thickness less than 10cm;otherwise the all section is accepted .from past experment ,the stat highway engineer know the 85%of all section constructed by the contructor comply with specification . however the relability of altrusonic thickness testing is only 75 ,so that there is a 25 percent chane of errorneous concolusion based on the determenation of thickness with ultrasonic . what is the probablity that a poorly constructed section is accepted on the base of ultrasonic test?

The chance or probability of getting accepted is 0.85; the chance of getting accepted even when bad is 0.25. So therefore the chance of being bad and getting selected can be solved using the conditional probability theorem given by:

P(A/B)= P(AnB)/P(B).

Going by this the answer is: 0.25 x 0.85= 0.2125

solution

the possible out come of rolling die is =6 here in this case since it is rolled 3 our sample space is 6×6×6=216

we have asked to solve the probability of sum which will be atleast 5 this means 5 and more is possible. so that we have to search the possibilities of less than five to easy our work this will be like[111][112][121] = 3 out comes onlywso p(s`)=3/216 when p(s`) is probability of sum less than five or probability of sum greater than equal to five.

since the sum of p(s) and p(s`)=1

p(s)=1-p(s~)

1-3/216=213/216

what about 2,1,1?

Two cards are drawn at random from an ordinary deck of 52 card. Find the probability P that

(a) Both are spade

(b) One is a spade and one is heart

Ans:

(a) Probability of getting spade 1st time is 13/52 and Probability of getting spade 2nd time is is 12/51

Total Probability is 13*12/(52*51) = 156/2652

(b) Probability of getting spade is 13/52 and Probability of getting Heart is 12/51

Total probability is 13*13/(52*51) = 169/2652

Question ‘b’ says that one is a spade and one is a heart. Therefore the possibilities are ‘heart-spade’ and ‘spade-heart’.

The answer should be:

((13*13/52*51) + (13*13/52*51)) = 169/2652 + 169/2652 =338/2652 = 13/102

copying the solution offerred by @ diriba

solution

the possible out come of rolling die is =6 here in this case since it is rolled 3 our sample space is 6×6×6=216

we have asked to solve the probability of sum which will be atleast 5 this means 5 and more is possible. so that we have to search the possibilities of less than five to easy our work this will be like[111][112][121] = 3 out comes onlywso p(s`)=3/216 when p(s`) is probability of sum less than five or probability of sum greater than equal to five.

since the sum of p(s) and p(s`)=1

p(s)=1-p(s~)

1-3/216=213/216

The above solution is good but a little faulty because it considered only the possibility of obtaining a ‘1’ on the first die, it omitted the possibility of getting a ‘2’ on the first die i.e (using the same notation) [211], this is the fourth possible outcome.

Hence P(s)= 1- P(s’)

= 1-4/216

=212/216

=53/54

A bag contains blue and red balls. Two balls are drawn randomly without replacement. The

probability of selecting a blue and then a red ball is 0.2. The probability of selecting a blue ball in the first

draw is 0.5. What is the probability of drawing a red ball, given that the first ball drawn was blue? Solution please

Lets assume probability of picking a red ball is X.

The probability of selecting a blue ball and then a red ball,

P(B)*P(R)=.2

.5*X=.2

x=.5/.2

x=.4

P(R|B)=P(R and B)/P(B)

=0.2/0.5=0.4

The probability of snow tomorrow is 0.6. And the probability that it will bi colder is 0.7. The probability that it will not snow and not bi colder is 0.1 .What is probability that it will not snow if it is colder tomorrow?

please solve it … and tell me answer.. thanks …

A= event for it will snow tomorrow

B= event for it will be cold

a= event for it will not snow

b= event for it will not be cold

P(A) = 0.6; P(B)= 0.7; P(a n b)=0.1

P(A)+P(B)-P(AnB)+P(a n b)=1

Inserting values, you’ll have

P(AnB)=0.4.

P(Bna)= P(B)-P(BnA)

This will give you P(Bna)= 0.3.

By conditional theorem,

P(a/B)=P(anB)/P(B)

This will give you 3/7 as the answer.

pl first draw a tree diagram:

P(S)=0.6, P(N.S.)=0.4

P(Cold)=0.7, P(Not Cold)=0.3

P(No S + Not Cold)= 0.4*0.3=0.12=0.1

P(No S + Cold)=0.4*0.7=0.28

in a class 10 boys and 5 girls .three students are selected random one after the other.find the probability that

1)first two are boys and third is girl

2)first and third is of same gender and third is of opposite gender

please help me in solving this

A) 10/15*9/14*5/13

B) 1st case:

1st & 2nd are boys & 3rd is girl

10/15*9/14*5/13

2nd case:

1st & 2nd are girls & 3rd is boy

5/15*4/14*10/13

n(b)=10

n(g)=5

n(t)=15

(a) p(first two are boys and third is a girl)

p(B,B,G)

10/15*9/14*5/13=

5/1*3/7*1/13=15/91

(b) p(first and third is of the same gender and second is opposite)

P(B,G,B) or p(G,B,G)

(10/15*5/14*9/13)+(5/15*10/14*4/13)=

(450/2730)+(200/2730)=

450+200/2730=

650/2730=

5/21

There are three boxes, one of which contains a prize. A contestant is given two chances, such that if he chooses the wrong box in the first round, that box is removed from the selection and he then chooses between the two remaining boxes.

1. What is the probability that the contestant wins?

2. Does the contestant’s probability of winning increases on the second round?

Hi Lei,

It’s a Monty Hall problem. You can google it.

As for your question,

As the first box chosen if found empty is removed and you HAVE/Switch to pick from other two, the P(W) = 2/3.

Above answer can be explained as Prob. of winning on first box + Prob. of choosing wrong * Prob. of Choosing right between the two => 1/3+2/3*1/2 => 2/3

The answer to the second: Yes probability increases as its a 50% chance to win as 1 wrong box is eliminated.

1) 10C2*5C1/15C3?

2) (10C1*5C1*9C1/15C3) + (5C1*10C1*4C1/15C3)?

Plz solve it

XYZ company wants to start a food outlet in pakistan. There is a 40% and 60% chance of stating in hyderabad and karachi respectively. If he start the outlet in hyderabad there is 30% chance that it will be in saddar and 70% chance that it will be in defence area. If they start the outlet in karachi there is 50% chance that it will be in defence, 30% in clifton and 20% in pechs. Determine probability of starting the outlet in: (a) saddar (b) defence area of any city (c) clifton given that the outlet is started in karachi

a) P(H,S) = 40% x 30% =0.4 x 0.3 = 0.12 = 12%

b) P(H, D) + P(K, D) = 40% x 70% + 60% x 50% = 0.4 x 0.7 + 0.6 x 0.5 = 0.28 + 0.3 = 0.58 = 58%

c) P(C|K) = 30%

In maternity clinic the probability of new born was females is 55%=0.55

So,the probabilitt of the next three deliveries are females is 0.55×0.55×0.55=0.166 or 16.6%

Nah, that’s not correct. Births are independent events. It doesn’t matter what the gender of previous births were. The odds of the next birth being female is still 50%. The probability of three females in a row is simply (0.5)^3 = 1/8 = 0.125

a) 15*(0.2)^4*(0.8)^2 = 0.01536

b) 0.01536 + 6*(0.2)^5*(0.8)^1 + 1*(0.2)^6*(0.8)^0 = 0.01696

c) (0.8)^6 = 0.262144

hopefully you haven’t been waiting over a year for a response 🙂 Here you go though…

A: (2/3)^ 3 = 8/27

B: 1 * 2/3 * 1/3 = 2/9

C: 1 * 1/3 * 1/3 = 1/9