Dependent and Independent Events – Probability

Probability theory is an important topic for those who study mathematics in higher classes. For example, the weather forecast in some areas says that there is a fifty percent probability that it will rain today. The probability is a chance of some event happening. 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:

• Obtaining a tail in a toss of a coin may be called an event.
• Getting a 4 on a roll of a die is said to be an event.
• Drawing a king from the deck of cards is also an event.
• 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 are 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 a 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 = Number of favorable outcomes/ Total number of outcomes

Example: The probability of getting a 4 when a die is tossed.

Solution:

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 a compound event. 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.

Example: The probability of rolling an odd number on a die, then tossing a tail on a coin.

Solution:

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 the 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.

Solved Examples on Dependent and Independent Events 

Question 1: An instructor has a question bank with 300 easy T/F, 200 Difficult T/F, 500 easy MCQ, and 400 difficult MCQ. If a question is selected randomly from the question bank, What is the probability that it is an easy question given that it is an 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 apples, 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 is 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)
• 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 consists 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 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 on the white die is 5.

b) The score on the white die is 1 and the 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.