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Data Handling

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Data Handling


Practical implementation of mathematics can be found out in a wide range of daily routine lives to a great extent. Several prominent concepts in mathematics find significant usage in the daily world. One of such concepts is the Probability that we tend to use in our day to day lives.

Probability could be understood as a numerical description of the likelihood of occurrence of an event. It is a number that lies between a range of 0 and 1. Here, 0 directly defines impossibility of occurrence of event and 1 tend to define certain of the event. For instance, events such as rolling a dice, pulling out a card from a deck of cards or pulling a coloured ball from a bag are all uncertain, and we are not sure what the outcome would be. It could be understood that the outcome of such events in random and can be termed as Random Variable (RV).

Important Terminology

Mutually Exclusive Events

Two events are said to mutually exclusive if they cannot occur simultaneously. In other simpler words, the occurrence of an event would preclude the occurrence of all other associated events.

For instance, if a ball is red, it cannot be white at the same time. If a person is dead, he/she cannot be alive at the same time. If a coin is tossed, either tails or heads will occur as both of them cannot occur simultaneously.

Dependent and Independent Events

Events are said to be independent primarily when the occurrence of one event does not impact the another. It directly indicates that one trial will not describe anything about other trials.

For example, tossing a coin is an independent event. This is because if a coin is tossed, one trial will not be impacted by the other.

However, dependent events could be understood as an event where the occurrence of one trial would have an impact on the occurrence or non-occurrence of another event. For example, if a card is drawn from a deck of playing cards and is not replaced, then the Probability of the second trial will be impacted.

Equally Likely Event

Two events could be defined as equally likely when both of them have an equivalent chance of occurrence. In simpler words, two events can only be termed as equally likely if one of the events does not occur more often in comparison to others.

For example, if an unbiased dice is thrown, each of six faces could be expected to occur in an equivalent number of times. In case, the dice is biased; then the face can not be expected to occur in equal number.


It could be defined any situation or phenomenon such as rolling of dice, tossing of coins, pulling out cards from a deck of, etc.


The outcome is defined as the result of an event after completion of the experiment. For instance, side of coin after the toss, the appearance of a number on dice, etc. could be considered as potential outcomes of experiments.


The event could be defined as a combination of all probable and plausible outcomes associated with an experiment. Obtaining tail or head on tossing a coin, obtaining odd or even number after rolling dice, etc. are all considered as events.

Sample Space

It could be defined as the set that comprises of all possible outcomes or result.

Formulae and Notations

Probability of occurrence of event A will be denoted as P(A). Similarly, the Probability of occurrence of event B will be denoted as P(B).

P(A) = n(E) / n(S)

Here, n(E) = Number of Events and n(S) = Sample Space

To denote non-occurrence of event A, symbol A` is used and the notation would be P(A`).

P (A. A`) = 0

P (A.B) + P (A`. B`) = 1

P(A`B) = P(B) – P (A.B)

P(AB`) = P(A) – P (A.B)

Probability of occurrence of Event A or B is denoted by P(A∪B). The symbol ∪ is known as a union.

P(AB) = P(A) + P(B) – P(A∩B)

Probability of occurrence of Event A and B is denoted P(A∩B). The ∩ symbol is known as the intersection.

P(A∩B) = P(A). P(B)
















For obtaining a better grasp on the concept of Probability, let’s practice some problems that will help in enhancing the overall clarity.

Q1. One of your friends rolls two dices simultaneously. Find out the Probability of getting a sum of 6after rolling two dices.

  1. 5/36
  2. 4/36
  3. 6/36
  4. 2/36

Answer – (a)

Solution – Let’s denote the experiment with symbol – A. For solving this question, we have to find out the sample space of the experiment. Sample space for the proposed experiment has been given below.

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)


Favourable number of outcomes = {(1,5), (2,4), (3,3), (4,2), (5,1)}

∴ n(E) = 5. Further, the sample space comprises of 36 elements, n(S) = 36.

Hence, P(A) = n(E) / n(S) = 5/36.

It denotes that 5 out of 36 times, we will obtain a pair whose sum will be equivalent to 6.


Q2. Three unbiased coins are thrown simultaneously. Find out the overall Probability of occurrence of two heads and one tail.

  1. 1/7
  2. 2/9
  3. 1/8
  4. 3/8

Answer – (d)

Solution – Before solving this question, we need to find out the exact sample space for the experiment.

Sample Space –

Toss 1 Toss 2 Toss 3


We can observe that there isa total of 8 outcomes associated with the experiment. However, we need to find out plausible outcomes where two heads and one tail is occurring –

Favourable Outcomes – {(H, H, T), (H, T, H), (T, H, H)}

Out of 8 outcomes, only 3 outcomes are favourable. Hence, the Probability of an event is 3/8.

Q3. Samantha drew one card from a pack of well-shuffled cards. What is the Probability that she drew a queen?

  1. 1/14
  2. 4/52
  3. 1/12
  4. 4/26

Solution – A deck of well-shuffled card comprises of 52 cards. There are a total of 4 queens in the entire deck of card.

Hence, the total sample space is 52, and favourable outcomes are 4. Hence, the answer to the problem is 4/52 or 1/13.

Q4. Andrew drew one card from a deck of cards. Find out the Probability that he drew either heart or a club?

  1. 13/52
  2. ½
  3. 67%
  4. 33%

Answer – (b)

Solution – In a pack of a well-shuffled deck, a total of 52 cards arethere which are of four different suits – hearts, clubs, spades, and diamonds. Each suite has 13 cards.

Hearts and clubs will have a total of 26 cards. Hence, favourable outcomes are 26, and the size of the sample is 52. Thus, the Probability will be 26/52 or 1/2. Therefore, the answer to the problem is (b).

Q5. A box comprises of 6 green, 7 blue, and 8 red balls. One of your friends picks up a ball randomly. Find out that the ball is neither green nor red.

  1. 2/3
  2. 1/3
  3. 4/3
  4. 2/5

Answer – (b)

Solution – Total number of balls in the bag are – (6 +7 + 8) = 21 balls.

Assume that E = event where neither green nor red ball is fetched = event where only blue ball is drawn. Hence, n(E) = 7.

∴ P(E) = n(E) / n(S) = 7/21 = 1/3. Thus, correct answer to the problem is (b).

Q7. Which of the following is true in the context of Probability?

  1. Probability of an event lies an in-between range of 0 and 1, where 0 and 1 are not included
  2. Probability of an event is always greater than 1
  3. Probability of an event can only lie within the range of 0 and 1, where both limits are included
  4. Probability of an event can be negative under some circumstances

Answer – (c)

Solution –As per the concept of Probability, an event can either happen or not happen. Using this definition, we can say that the Probability of an event will surely lie between the range of 0 and 1. Further, some of the events will never occur under any circumstances whereas some events will always be true. Thus, the Probability will comprise of both limits. Also, 0 is the minimum limit, and 1 is the maximum limit of Probability.

Hence, all such factors define that only option (c) is correct, and other options are incorrect.

Q8. Find out the Probability of occurrence of event A or event B, if the Probability of A is 0.5, the Probability of B is 0.7, and the Probability of event A and Event B is 0.3.

  1. 2
  2. 5
  3. 7
  4. 9

Answer – (d)

Solution –For finding an appropriate solution to this problem, we will have to use the rudimentary formula, i.e. P(AB) = P(A) + P(B) – P(A∩B)

P(A) = 0.5, P(B) = 0.7, and P(A∩B) = 0.3.

By putting these values in the formula, we can find out the Probability of occurrence of event A or B.

P(AB) = 0.5 + 0.7 – 0.3 = 0.9

Q9. A carton comprises of 10 eggs out of which 3 are broken. Chris takes 2 eggs from the carton and prepares omelette. Later, he took another egg from the carton to prepare another omelette. Find out the overall Probability that this egg was not broken.

  1. 1/8
  2. 5/8
  3. 2/10
  4. 3/10

Answer – (b)

Solution – Initially, Chris had 10 eggs in total out of which 3 were broken. Only 7 eggs were fine and unbroken. However, he took out 2 unbroken eggs. Hence, 8 total eggs remained in the carton. All broken eggs are still in the carton as Chris has not encountered any broken egg. Thus, as per current scenario, the bag comprises of 8 balls out of which 3 are broken, and 5 are unbroken.

Hence, favourable outcomes are 5 and total outcomes are 8.

By using the basic formula of Probability, we can deduce that answer to the problem is 5/8.

Q10. There are 20 tickets which are numbered from 1 to 20, and a random ticket was drawn at once. Find out the Probability that the ticketwas drawn possess a number that is either a multiple of 3 or 5?

  1. 2/3
  2. 3/5
  3. 10/20
  4. 1/5

Solution – Sample Space of the experiment will be – {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}.

A (Event for getting multiple of 3 or 5) = {3, 5, 6, 8, 9,10, 12, 15, 18, 20}

P (A) = n(E)/n(S) = 10/20. Hence, 10 out of 20 tickets have a number that has a number which is either a multiple of 3 or 5. Thus, correct solution of the problem is 10/20.∴ Option (c)




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