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Probablity and Its Application

Probability is the measure of uncertainty. We turn our attention to one of the problems that is widely held responsible for the development of the theory of probability, namely, that of throwing a dice. As all of us know, a dice is a well-balanced cube with six faces marked with numbers from 1 to 6, one number on one face.
 
Let’s throw a dice once. What are the possible outcomes? Clearly, a dice can fall with any of its faces uppermost. The number on each of the faces is, therefore a possible outcome and all the outcomes are equally probable. Hence, it is as likely to show up a number, say 2, as any other number 1, 3, 4, 5, or 6.
 
Since there are six equally likely outcomes in a single throw of a dice and there is only one way of getting a particular outcome (say 2). Therefore, the chance of the number 2 coming up on the dice is 1 by 6. In other words, the same phenomenon is known as probability of getting 2 in a single throw of dice is 1/6.
 
We write this as P(2) = 1/6
 
Similarly, when an ordinary coin is tossed, it may show up head (H) or tail (T). Hence, the probability of getting a head in a single toss of a coin is given by
 
P(H) = ½

Before we define the process to finding out the probability, it is essential to understand the various terms associated with probability.

Trial and Elementary Events

When we repeat a random experiment under identical conditions, then the experiment is known as a trial and the possible outcomes of the experiment are known as elementary events.
 
For example,
  • tossing of a coin is a trial and getting head or tail is an elementary event.
  • throwing of a dice is a trial and getting 5 on its upper face is an elementary event.

Compound Events

Events obtained by combining two or more elementary events are known as the compound events. A compound event is said to occur if one of the elementary events associated with it occurs.

Exhaustive Number of Cases

The total number of possible outcomes of a random experiment in a trial is known as the exhaustive number of cases.
 
For example, in throwing of a dice the exhaustive number of cases is 6, since any one of the six face marked with 1, 2, 3, 4, 5, 6 may appear on its upper face.

Mutually Exclusive Events

Events are said to be mutually exclusive if the occurrence of any one of them prevents the occurrence of all the others, i.e., if no two or more than two events can occur simultaneously in the same trial.

Equally Likely Events

Events are equally likely if there is no reason for an event to occur in preference to any other event.
 
For example, while throwing a dice, chances of occurring of Head or Tail are an equally likely event.

Favourable Number of Cases

The number of cases favourable to an event in a trial is the number of elementary events such that if any one of them occurs, we say that the event happens.
 
In other words, the number of cases favourable to an event in a trial is the total number of elementary events such that the occurrence of any one of them ensures the happening of the event.
 
For example, in throwing of a dice, the number of cases favourable to the appearance of a prime number is 3 viz., 2, 3 and 5.

Independent Events

Events are said to be independent if the happening (or non-happening) of one event is not affected by the happening (or non-happening) of others.
 
For example, if two cards are drawn from a well-shuffled pack of 52 cards one after the other with replacement, then getting an ace in the first draw is independent of getting a jack in the second draw. But, if the first card drawn in the first draw is not replaced, then the second draw is dependent on the first draw.

Sample Space

This is the most important factor in probability and is defined as the set of all possible outcomes of a random experiment associated with it.
 
These examples suggest the following definition of probability (assuming that outcomes are equally likely).
 
Now, Probability of an event E, written as P(E), is defined as
Description: 1500.png 

Points to Remember
  1. 0 ≤ P(E) ≤ 1
  2. P(E) + P’(E) = 1
Example-1
In a single throw of two die, what is the probability that the sum on the top face of both the die will be more than 9?
Solution
When two die are thrown, sum of the numbers appearing on the faces can be anything from 2 to 12. To find the number of favourable cases we will be required to find the cases in which the sum is more than 9.
 
Following are the cases—(5, 5), (6, 4), (4, 6), (6, 5), (5, 6), (6, 6)
 
So, the total number of favourable cases = 6
 
The total number of possible outcomes = 6 × 6 = 36
 
Hence probability = 6/36 = 1/6
 

Example-2
Six dice are thrown simultaneously. Find the probability that all of them show the same face.
Solution
Sample space of throwing six dice = 6 × 6 × 6 × 6 × 6 × 6 = 66.
All dice are showing the same face implies that we are getting same number on the entire six dice. The number of ways for which is 6C1.
Hence, the required probability = Description: 1509.png 
 
 
Example-3
In the above question, find the probability that all of them show a different face.
Solution
The total number of ways in which all the dice show different numbers on their top-faces is the same as the number of arrangement of 6 numbers 1, 2, 3, 4, 5, 6 by taking all at a time.
 
So, the number of favourable cases = 6!
Hence, the required probability = Description: 1518.png
 

Example-4
Five persons enter a lift on the ground floor of an 8-floor apartment. Assuming that each one of them independently and with equal probability can leave the lift at any floor beginning with the first. What is the probability that all the five persons are leaving the lift at different floors?
Solution
Apart from the ground floor, there are 7 floors.
 
A person can leave the lift at any of the seven floors. Hence the total number of ways in which each of the five persons can leave the lift at any of the 7 floors = 75.
 
So, the sample space = 75
 
Five persons can leave the lift at five different floor = 7P5 ways.
 
So, the favourable number of ways = 7P5.
 
Hence, the required probability = Description: 1527.png
 

Example-5
If you have 3 tickets of a lottery for which 10 tickets were sold and 5 prizes are to be given, the probability that you will win at least one prize is:(JMET 2005)
  1. Description: 1536.png
  2. Description: 1545.png
  3. Description: 1554.png
  4. Description: 1563.png 
Solution
Probability that you will win at least one prize = 1 – probability that you will not win any prize.
Description: 1572.png
 
Odds in Favour and Odds Against
 
Odds in Favour = Description: 1581.png
 
Odds in Against = Description: 1590.png
 
Understanding And/Or To understand the role played by And/Or in our calculation, let us take the example of throwing an unbiased dice. Let A and B be the two events associated with it such that
A = getting an even number, B = getting a multiple of 3.
 
Then A = {2, 4, 6}, and B = {3, 6}.
 
We now define a new event “A or B” which occurs if A or B or both occur i.e., at least one of A or B occurs. Clearly the event “A or B” occurs if the outcome is any one of the outcomes {2,3,4,6}. Thus, the event “A or B” is represented by the subset A U B.
 
Similarly, “A and B” means occurrence of both A and B which is possible if the outcome is {6}.
 
Hence, it is represented by the subset A  B.





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