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Dependent and Independent Events – Probability
• Last Updated : 18 Mar, 2021

Probability theory is an important topic for those who study mathematics in higher classes. For example, Weather forecast of some areas says that there is a fifty percent probability that it will rain today. The probability is a chance of some event to happen. The term “event” actually means one or more outcomes. The event means the outcome which is able to occur. Total events are defined as all the outcomes which may occur relevant to the experiment asked in the question. Also, the events of interest are known as favorable events.

For example:

i) Obtaining a tail in a toss of a coin may be called an event.

ii) Getting a 4 on a roll of a die is said to be an event.

iii) Drawing a king from the deck of cards is also an event.

(iv) Getting a sum of 7 on the roll of a pair of dice is an event.

Sure and Unsure Events:

An event whose chances of happening is 100 % is called a sure event. The probability of such an event is 1. In a sure event, one is likely to get the desired output in the whole sample experiment.
On the other hand, when there are no chances of an event happening, the probability of such an event is likely to be zero. This is said to be an impossible event.

On the basis of quality events, these are classified into three types which are as follows:

• Independent Events
• Dependent Events
• Mutually-Exclusive Events

For better understanding of dependent and independent events, lets us first understand the simple and compound events

### Simple Event

An event that has a single point of the sample space is known as a simple event in probability.

Probability of an event occurring = No. of favorable outcomes/Total no. of outcomes

For example: the probability of getting a 4 when a die is tossed.

Here Sample Space = {1, 2, 3, 4, 5, 6}

If E be the event of getting a 4 when a die is tossed. E = {4}

P(E) = 1/6

In the case of a simple event, the numerator (number of favorable outcomes) will be 1.

### Compound Event

If an event has more than one sample point, it is termed as a compound event. The compound events are a little more complex than simple events. These events involve the probability of more than one event occurring together. The total probability of all the outcomes of a compound event is equal to 1.

To calculate probability, the following equation is used:

First, we find the probability of each event occurring. Then we will multiply these probabilities together. In the case of a compound event, the numerator (number of favorable outcomes) will be greater than 1.

For example: the probability of rolling an odd number on a die, then tossing a tail on a coin

Here P(odd number) = 3/6 where favorable outcomes are  {1,3,5}

P(tail) = 1/2

Hence, required probability = (3/6)×(1/2) = 1/4

Now Let’s come to Dependent and Independent events:

### Dependent Events

Dependent events are those  events that  are affected by the outcomes of events  that had already occurred previously. i.e. Two or more events that depend on one another are known as dependent events. If one event is by chance changed, then another is likely to differ.

Thus, If whether one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent.

For example:

1. Let’s say three cards are to be drawn from a pack of cards. Then the probability of getting a king is highest when the first card is drawn, while the probability of getting a king would be less when the second card is drawn. In the draw of the third card, this probability would be dependent upon the outcomes of the previous two cards. We can say that after drawing one card, there will be fewer cards available in the deck, therefore the probabilities tend to change.

2. A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. What is the probability that the first card chosen is a king and the second card chosen is a queen?

Probabilities:

P(king on first pick)= 4 /52

P(queen on 2nd pick given king on 1st pick) = 4 /51

P(king and queen) = (4/52 × 4/51) = 16/2652 = 4 /663

It involved two compounds, dependent events. The probability of choosing a queen on the second pick given that a king was chosen on the first pick is called a conditional probability

When the occurrence of one event affects the occurrence of another subsequent event, the two events are dependent events. The concept of dependent events gives rise to the concept of conditional probability.

### Conditional Probability Formula

If the probability of events A and B are P(A) and P(B) respectively then the conditional probability of B such that A has already occurred is P(A/B).

Given, P(A)>0,P(A/B) = P(A ∩ B)/P(A) or P(B ∩ A)/P(A)

P(A)<0 means A is an impossible event. In P(A ∩ B) the intersection denotes a compound probability.

### Sample Problems

Question 1: An instructor has a question bank  with 300 easy T/F, 200 Difficult T/F, 500 easy MCQ, 400 difficult MCQ. If a question is selected randomly from question bank, What is the probability that it is easy question given that it is a MCQ.

Solution:

Let,

P(easy)= (300+500)/1400 = 800/1400 = 4/7

P(MCQ)= (400+500)/1400 = 900/1400 = 9/14

P(easy ∩ MCQ)= (500)/1400 =5/14

P(easy/MCQ) = P(easy ∩ MCQ)

= (5/14)/(9/14) =5/9

Question 2: In a shipment of 20 apple, 3 are rotten. 3 apples are randomly selected. What is the probability that all three are rotten if the first and second are not replaced ?

Solution:

Probabilities: P(3 rotten) = (3/20 ×  2/19 × 1/18)= 6/6840 = 1/1140

Question 3: John has to select two students from a class of 10 girls and 15 boys. What is the probability that both students chosen are boys?

Solution:

Total number of students = 10 + 15 =25

Probability of choosing the first boy, say P(Boy 1) = 15/25

P(Boy 2|Boy 1) = 14/24

Now,

P(Boy 1 and Boy 2) = P(Boy 1) and P(Boy 2|Boy 1)

= (15/25)×(14/24) = 7/20

### Independent events

Independent events are those events whose occurrence is not dependent on any other event. If the probability of occurrence of an event A is not affected by the occurrence of another event B, then A and B are said to be independent events.

Examples:

• Tossing a coin.

Here, Sample Space S =  {H, T}  and both H and T are independent events.

• Rolling a die.

Sample Space S = {1, 2, 3, 4, 5, 6}, all of these events are independent too.

Both of the above examples are simple events. Even compound events can be independent events. For example:

• Tossing a coin and rolling a die.

Sample space S = {(1,H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.

These events are independent because only one can occur at a time.

Consider an example of rolling a die. If A is the event ‘the number appearing is greater than 3’ and B be the event ‘the number appearing is a multiple of 3’, then

P(A)= 3/6 = 1/2 here favorable outcomes are {4,5,6}

P(B) = 2/6 = 1/3 here favorable outcomes are {3,6}

Also, A and B is the event ‘the number appearing is odd and a multiple of 3’ so that  P(A ∩ B) = 1/6

P(A│B) = P(A ∩ B)/ P(B)

= (1/6)/(1/3) = 1/2

P(A) = P(A│B) = 1/2, which implies that the occurrence of event B has not affected the probability of occurrence of the event A.
If A and B are independent events, then P(A│B) = P(A)

Using Multiplication rule of probability, P(A ∩ B) = P(B). P(A│B)

P(A ∩ B) = P(B). P(A)

Note: A and B are two events associated with the same random experiment, then A and B are known as independent events if P(A ∩ B) = P(B).P(A)

Note: We can calculate the probability of two or more Independent events by multiplying

### What are Mutually Exclusive Events?

Two events A and B are said to be mutually exclusive events if they cannot occur at the same time. Mutually exclusive events never have an outcome in common.

Question 1: A multiple choice test consist of two problems. Problem1 have 5 option and Problem2  have 4 options. Each problem has only one correct answer. What is the probability of  randomly guessing the correct answer to both the problems.

Solution:

Here, the probability of correct answer of Problem1 = P(A)  and the probability of correct answer of Problem2 =P(A) are independent events. Thus

The probability of correct answer of Problem1 and Problem2 both = P(A ∩ B) =P(A). P(B)

=(1/5)*(1/4) = 1/20

Question 2: If a die is thrown twice, find the probability of getting two 3’s.

Solution:

P(getting 3 on first throw)=1/6

P(getting 3 on second throw)=1/6

P(two 3’s)=(1/6)*(1/6) =1/36

Question 3: Two fair dice, one colored white and one colored black, are thrown. Find the probability that:

a) the score on the black die is 3 and white die is 5.

b) the score on the white die is 1 and black die is odd.

Solution:

a) Probability the black die shows 3 and white die 5 =   (1/6)*(1/6) = 1/36

b) Probability the white die shows 1 and black die shows an odd number = (1/6)*(3/6) =1/12

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