# Probability | Theory, solved examples and practice questions-Titik Letak-Titik Letak

# Probability | Theory, solved examples and practice questions:

# Meaning of the 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 2 pens are drawn at random from the pack, replaced and then another pen is drawn. 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 = 8/81

#### 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}) = 1For 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, by randomly choosing. What is the probability of 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

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