# How can I solve any question on finding the probability of an event in a job interview / written test? (Part 2)

**Properties of probability**

i) ** 0 ≤ P(A) ≤ 1**.

ii) ** P(ɸ) = 0, P(S) = 1**.

iii) **For any two events A**

**and**,

*B***, where**

*P(AUB) = P(A) + P(B) – P(A∩B)**AUB*is read as ‘

*A*

*union*

*B’*or in common parlance ‘

*A*

*or*

*B’*, and

*A∩B*is read as ‘

*A*

*intersection*

*B’*, or in common parlance ‘

*A*

*and*

*B’*.

iv) *P(A*^{c}** ) = 1 – P(A)**, where

*A*denotes the complement of event

^{c}*A*.

v)** If A and B are mutually exclusive events, then**

**. A ‘mutually exclusive event’ means that if event**

*P(A∩B) = 0**A*occurs, then event

*B*cannot occur. In other words, there is no common element between

*A*and

*B*, i.e.,

*A∩B = ɸ*, the null event, and hence

*P(A∩B) = 0*[property ii]. For example, consider the events

*A*= head (

*H*), and

*B*= tail (

*T*), in tossing of a coin. They are mutually exclusive because if you get the head, you cannot get the tail in any single toss of a coin, or, in other words, they cannot occur together.

vi) **If A**

**and**

*B***are independent events, then**

**. ‘Independent events’ are those where the outcome of one event does not affect the outcome of another event. For example, if you are tossing two coins, the outcome of the first coin will not affect the outcome of the second coin.**

*P(A∩B) = P(A)*P(B)*Any question regarding the probability of an event can be classified into two categories:

Category 1: Probability of each sample point is not given in the question.

Category 2: Probability of each sample point is given in the question.

**Category 1: **

All the questions under this category, you will be given the probability values of one or more events and you will be asked the probability value of a related event. You have to use the properties of probability to solve these questions. For example:

** ****Q:** If P(A) =0.3, P(B)= 0.4 and A and B are independent events, find P(AUB).

**Ans**: P(AUB) = P(A) + P(B) – P(A∩B) [property iii]*Or, P(AUB) = P(A) + P(B) – P(A)*P(B) [property vi]**Or, P(AUB) = 0.3 + 0.4 – (0.3)(0.4)**Or, P(AUB) = 0.7 – 0.12 = 0.58.*

**Q: **If A and B are two events such that P(A) =0.3, P(B)= 0.4 and the probability that at least one of them occurs is 0.8, then what is the probability of both of them to occur?

**Ans:** Given P(A) =0.3, P(B)= 0.4, and P(AUB) = 0.8. *P(AUB) = P(A) + P(B) – P(A∩B) [property iii]**Or, P(A∩B) = P(AUB) – P(A) – P(B)**Or, P(A∩B) = 0.8 – 0.3 -0.4 = 0.1.*

**Category 2: **

Under this category, the probability of each sample point will be given. This case can be divided into two types, depending on the nature of the sample space.

**Type 1- Discrete type sample space:** Under this type, probability of each sample point will be given to you and you will be asked to find the probability of an event *A*. To solve this type of problem, you have to first identify all the sample points that consist *A*. Then *P(A)* is nothing but the sum of all individual probabilities of the corresponding sample points of *A*. If the probability of each sample point is not given to you, then assume that each sample point has equal probability.

**Q:** Suppose two coins are tossed together. What is the probability of observing

*a) both heads?**b) at least one head?*

**Ans**: Note that the sample space S of this experiment is {HH, TH, HT, TT}. Since the probability of each sample point is not given in the question, we assume they are equally probable, i.e., 0.25 in this case.

a) *Here, our favorable event A consists of only one sample point HH, where both heads occur. Hence, P(A) = P(HH) = 0.25. *

b) *Here, our favorable event A consists of three sample points HH, TH and HT, because these are the 3 out of 4 possible outcomes where at least one head occur. Therefore, P(A) = P(HH)+P(TH)+P(HT) = 0.75. *

**Q:** Suppose that the probability of occurring x accidents per year in a busy street in Mumbai is given by

*What is the probability that there is at least one accident in this year?*

**Ans: **Note that the number of accidents in this year can be any number between 0 to infinity. Here the sample space S = {0,1, 2, 3,…} and the probability value for each sample point are given. The favorable event (or the event of interest as asked for in the question) is A = { 1, 2, 3,…}. So,

**Type 2- Interval type sample space:** Under this type, a probability function will be given for an interval. This probability function is known as probability density function (pdf). You will be asked to find the probability of an event *A*, which is also an interval. To find the probability *P(A)*, you have to integrate the pdf under the interval *A*.

**Q: **Suppose the pdf of a process x is given by

*Find P(x>1). *

**Ans:** Here the sample space S is the interval (0, 2) and event A is the interval (1, 2). Hence,

Note: When you use R, Python, etc., the above calculations are done through the execution of simple commands, and you would not have to manually compute the above probabilities.

Thus, any question regarding finding the probability of an event fundamentally falls into any of these categories and can be solved accordingly. Some shortcuts may exist for certain problems. And in many problems, the fundamental information regarding the probability of each sample point is given under many jargons. Your job is to correctly identify it. Sometimes you may find it difficult to write the favorable event *A* in subset notation. I will suggest you to first understand the process *E* and write down corresponding the sample space *S*. Identifying the event *A* and finding *P(A)* will be easy then.