[Maths Class Notes] on Probability Questions Pdf for Exam

The word Probability refers to a chance or a possible outcome and talks about the possibility of a certain event that can occur. Mathematically, the term Probability is used to describe the likelihood of something that might happen or occur, and it is also considered to be the ability to understand as well as estimate the possibility of an outcome with different types of combinations. 

Types of Probability 

There are 3 main types of probabilities and they are as follows: 

1. Theoretical Probability: 

If you have access to the statistical data of a particular event, then you can precisely predict the said event. In statistics, the definition of Probability relies on whether an outcome is going to occur or not, i..e, the possibility of its occurrence. For example, if an individual wants to find out the theoretical Probability of getting a six on a rolling die, then they should start by determining the number of outcomes that are possible in that scenario. A die has6 numbers including 1, 2, 3, 4, 5, and 6. That is why the number of outcomes that are possible are also 6. Therefore, the chance of rolling the die and getting the number 6 is said to be 1:6. 

2. Experimental Probability: 

In statistics, unlike the definition of theoretical Probability, that of experimental Probability tends to include the number of trials as well. It is also referred to empirical Probability quite commonly as it is based on actual experiments and the recordings of the happenings of an event that happen to be very adequate. 

For example, an individual tosses a coin 20 times and out of those 20 times, they get tails for 10 times. That way, the experimental Probability of getting heads on the coin becomes 10:20. This particular calculation is specifically based on an experiment that has been carried out beforehand. 

Experimental Probability equals the number of all the outcomes of a particular event that are possible divided by the total number of trials. 

3. Axiomatic Probability: 

Axiomatic Probability refers to the theory of unifying Probability wherein there takes an application of a particular set of rules that were made by Kolomogorv. 

There are three main axioms and they are as follows: 

• Consider an event A. The Probability of said event A is always either going to be more than zero or equal to it, but it can never be less than zero. 

• If S is a sample space, then the Probability that a sample space will occur will always be 1, given the experiment is performed; then it is sure to get at least one of the said sample spaces. 

• For events that tend to be mutually exclusive, the Probability of either of those events happening is the sum total of the Probability of both of those events happening. 

Formula of Probability 

A particular event or experiment is said to have outcomes that are equally likely, when the possibility of the occurrence of each given outcome is the same in that particular event. 

So, on the basis of the experimental formula, it can be said that:

P(E)= the number of trials in which an event happened / total number of trials 

And on the basis of the theoretical formula, it can be said that: 

P(E)= the number of outcomes favourable to E / number of all possible outcomes of the experiment. 

Did You Know?

It was a mere gambler’s dispute in 1654 that led two of the most famous French mathematicians Blaise Pascal and Pierre de Fermat to come up with the mathematical theory of probability. 

The likelihood of an event to occur or the extent to which the event is probable is called probability. There are many events which cannot be predicted with total certainty, but we can only expect the chance of the event to occur. The likelihood of an event ranges from 0 to 1, where 0 means the event to be an impossible one, and one represents the event to be certain. If we have to calculate the probability of a single event to occur, we should know the total number of possible outcomes. Suppose, the probability of an event E happening is a number P(E), then:

0 ≤ P(E) ≤ 1

The likelihood of all events in a sample space adds up to 1.

Probability Formula

Suppose an event E occurs, then the probability of that event to occur P(E) is:

P(E) = The ratio of the number of favourable outcomes by the total number of outcomes.

The basic formula is shown above, but there are more formulas in different probability questions which arise due to different situations and events.

Example of a Probability Question

Suppose we toss two coins in the air, then four outcomes can happen, i.e. (H, H), (H, T), (T, H), (T, T). Now, if the probability of an event E containing two heads is asked, then:

P(E) = 1/4, since there is only one case where there are two heads, and there are in total of four outcomes.

Types of Events

There are two types of events:

Impossible Event: The events in which the probability is 0, such events are called impossible events.

Sure Event: When the probability of an event to occur is 1, that is the event certain to happen, such events are called sure events.

Probability Questions and Answers

These are some probability questions with its solutions:

1. Two coins are tossed 700 times, and the results are:

Two heads: 205 times

One head: 325 times

No head: 170 times

Find the probability of each event to occur.

Ans: Suppose that the events of getting no head, one head and two heads by E1, E2 and E3, respectively.

P(E1) = 170/700 = 0.24

P(E2) = 325/700 = 0.47

P(E3) = 205/700 = 0.29

Since, the sum of probabilities of all events of a random experiment is 1, therefore:

P(E1) + P(E2) + P(E3) = 0.24 + 0.47 + 0.29 = 1

2. If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, evaluate P(A/B).

Ans:

Here, P( A ), P( B ) and P(A∩B)  is given, we know that –

P(A|B) = (A∩B)/ P(B). Therefore:

(4/13) / (9/13) = 4/9

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Solved Examples

These are some solved examples of probability questions as follows:

1. Two dice are rolled, find the probability that the sum of the number of both the   dice is:

Equal to 2

Equal to 5

Less than 12

Ans:

S = {(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)}

Total number of possible outcomes, i.e. the sample space = 36

1) Let A be the event of the sum being equal to 2. Since, there is only one outcome where the sum is equal to 2, hence,

P(A) = ratio of n(A) and n(S) = n(A)/n(S) = 1 / 36 

2) Let B be the event of the sum being equal to 5. There are four possible outcome that give a sum equal to 5, they are:

B = {(1,4), (2,3), (3,2), (4,1)}

Hence, P(B) = n(B) / n(S) = 4 / 36 = 1 / 9

3) Let C be the event of the sum being less than 12. From the sample space, we can see all possible outcomes for event C, which give a sum less than 12. Like:

(1,1) or (1,6) or (2,6) or (5,6).

We can see that there are 35 outcomes in which the sum is less than 12. Hence,

P(C) = n(C) / n(S) = 35/36.

2. Calculating the probability of selecting a black card or a six from a deck of 52 cards.

Ans:

Let B be the event of selecting a black card.

We need to find out P(B or 6)

Selecting a black card P(B) = 26/52 = 1/2

Selecting a 6 P(6) = 4/52

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

We know that-

P(B ∪ 6) = P(B) + P(6) – P(B ∩ 6)

Therefore 

= 26/52 + 4/52 – 2/52

= 28/52

= 7/13.