Probability - A theoretical approach
Probabilities and likelihoods are everywhere around us, from weather forecasting to games, insurance or election polls. However, in the history of mathematics, probability is actually a very recent idea. While numbers and geometry were studied by ancient Greek mathematicians more than 2500 years ago, the concepts of probability only emerged in the 17th and 18th century.
According to legend, two of the greatest mathematicians,
To distract from the difficult mathematical theories they were discussing, they often played a simple game: they repeatedly tossed a coin – every heads was a point for Pascal and every tails was a point for Fermat. Whoever had fewer points after three coin tosses had to pay the bill.
One day, however, they get interrupted after the first coin toss and Fermat has to leave urgently. Later, they wonder who should pay the bill, or if there is a fair way to split it. The first coin landed heads (a point for Pascal), so maybe Fermat should pay everything. However, there is a small chance that Fermat could have still won if the
When we speak of a coin, we assume it to be ‘fair’, that is, it is symmetrical so that there is no reason for it to come down more often on one side than the other. We call this property of the coin as being ‘unbiased’. By the phrase ‘random toss’, we mean that the coin is allowed to fall freely without any bias or interference
Pascal and Fermat decided to write down all possible ways the game could have continued:
All four possible outcomes are equally likely, and Pascal wins in
Thus, they decided that Fermat should pay 3/4 of the bill and Pascal should pay 1/4.
Pascal and Fermat had discovered the first important equation of probability: if an experiment has multiple possible outcomes which are all equally likely, then:
Probability of an event =
In our example, the probability of Pascal winning the game is
Probability for winning in a dice game: For another example of equally likely outcomes, suppose we throw a die once. For us, a die will always mean a fair die. What are the possible outcomes?
This table shows all possible outcomes of throwing a die:
| X | 1 | 2 | 3 | 4 | 5 | 6 |
1 |
They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6. Let us try some:
Probability of getting 1 on the die =
Probability of (getting an even number on the die) =
=
Probability in drawing a specific colour ball: Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball without looking into the bag. What are the outcomes? Are the outcomes — a red ball and a blue ball equally likely?
Since there are 4 red balls and only one blue ball, you would agree that you are more likely to get a red ball than a blue ball.So, the outcomes (a red ball or a blue ball) are not equally likely.
However, the outcome of drawing a ball of any colour from the bag is equally likely. So, all experiments do not necessarily have equally likely outcomes.
Note: However, in this chapter, from now on, we will assume that all the experiments have equally likely outcomes.
We have already defined the experimental or empirical probability P(E) of an event E as:
P(E) =
The empirical interpretation of probability can be applied to every event associated with an experiment which can be repeated a large number of times. The requirement of repeating an experiment has some limitations, as it may be very expensive or unfeasible in many situations. Of course, it worked well in coin tossing or die throwing experiments.
But how about repeating the experiment of launching a satellite in order to compute the empirical probability of its failure during launching, or the repetition of the phenomenon of an earthquake to compute the empirical probability of a multistoreyed building getting destroyed in an earthquake?
In experiments where we are prepared to make certain assumptions, the repetition of an experiment can be avoided, as the assumptions help in directly calculating the exact (theoretical) probability.
The assumption of equally likely outcomes (which is valid in many experiments, as in the two examples above, of a coin and of a die) is one such assumption that leads us to the following definition of probability of an event.
The theoretical probability (also called classical probability) of an event E, written as P(E), is defined as:
P(E) =
,where we assume that the outcomes of the experiment are equally likely.
Let us find the probability for some of the events associated with experiments where the equally likely assumption holds.
Probability theory had its origin in the 16th century when an Italian physician and mathematician J.Cardan wrote the first book on the subject, The Book on Games of Chance. Since its inception, the study of probability has attracted the attention of great mathematicians.
Laplace’s Theorie Analytique des Probabilités, 1812, is considered to be the greatest contribution by a single person to the theory of probability. In recent years, probability has been used extensively in many areas such as biology, economics, genetics, physics, sociology etc.
Example 1 : Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
- In the experiment of tossing a coin once, the number of possible outcomes is
— Head (H) and Tail (T) - Let E be the event ‘getting a head’. Thus, P(E) = P (head) =
- Similarly, if F is the event ‘getting a tail’, then P(F) = P(tail) =
- We have found the answers
Proceeding to the next example
Example 2 : A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the (i) yellow ball? (ii) red ball? (iii) blue ball?
- Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.
- Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken out is blue’, and R be the event ‘the ball taken out is red’. Now, the number of possible outcomes =
. - So, P(Y) =
Number of yellow balls taken out by Kritika Total number of outcomes - P(R) =
Number of red balls taken out by Kritika Total number of outcomes - P(B) =
Number of blue balls taken out by Kritika Total number of outcomes - We have found the answers
Remarks :
- An event having only one outcome of the experiment is called an elementary event.
In Example 1, both the events E and F are elementary events i.e. : P(E) + P(F) = 1.
Similarly, in Example 2, all the three events, Y, B and R are elementary events i.e.: P(Y) + P(R) + P(B) = 1.
- Observe that the sum of the probabilities of all the elementary events of an experiment is 1. This is true in general also.
Moving on to the next example.
Example 3 : Suppose we throw a die once. (i) What is the probability of getting a number greater than 4 ? (ii) What is the probability of getting a number less than or equal to 4 ?
- Let
E be the event ‘getting a number greater than 4’. The number of possible outcomes is - That is: 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. Therefore, the number of outcomes favourable to E is
. - So, P(E) = P(number greater than 4) =
2 6 - Let F be the event ‘getting a number less than or equal to 4’. So, the number of outcomes favourable to F is
. - Outcomes favourable to the event F are 1, 2, 3, 4. Therefore, P(F) =
4 6 - We have found the answers
Remarks : From Example 1, we note that
P(E) + P(F) =
where E is the event ‘getting a head’ and F is the event ‘getting a tail’.
From (i) and (ii) of Example 3, we also get
P(E) + P(F) =
where E is the event ‘getting a number > 4’ and F is the event ‘getting a number ≤ 4’.
Note that getting a number not greater than 4 is same as getting a number less than or equal to 4, and vice versa.
In (1) and (2) above, is F the same as ‘not E’?
We denote the event ‘not E’ by
So, P(E) + P(not E) = 1
i.e. P(E) + P(
In general, it is true that for an event E:
P(
The event E , representing ‘not E’, is called the complement of the event E. We also say that E and
Before proceeding further, let us try to find the answers to the following questions:
(i) What is the probability of getting a number 8 in a single throw of a die?
To get number 8 in a single throw, we need to have a number 8 on the die which isn't possible. That is, the probability of an event which is impossible to occur is
Such an event is called an impossible event.
(ii) What is the probability of getting a number less than 7 in a single throw of a die?
Since every face of a die is marked with a number less than 7, it is sure that we will always get a number less than 7 when it is thrown once. So, the number of favourable outcomes is the same as the number of all possible outcomes, which is 6.
Therefore, P(E) = P(getting a number less than 7) =
For an event, where the probability was 1(getting a number less than 7) it is known as an sure event or certain event.
Note : From the definition of the probability P(E), we see that the numerator (number of outcomes favourable to the event E) is always less than or equal to the denominator (the number of all possible outcomes). Therefore,
0 ≤ P(E) ≤ 1
Probability related to playing cards: Now, let us take an example related to
The cards in each suit are ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2.
Kings, queens and jacks are called face cards.
Example 4 : One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will:
(i) be an ace, (ii) not be an ace.
- Well shuffled cards mean equally likely outcomes.
- There are
aces in a deck. Let E be the event ‘the card is an ace’. Thus, number of outcomes favourable to E = while number of possible outcomes = - Therefore, P(E) =
4 52 - Let F be the event ‘card drawn is not an ace’. The number of outcomes favourable to the event F =
– = - Therefore, P(F) =
= - We have found the answers
Remark : Note that F is nothing but
Example 5 : Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
- Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
- The probability of Sangeeta’s winning = P(S) =
(given) - The probability of Reshma’s winning = P(R) = 1 – P(S) =
- We have found the answers
Moving onto next example.
Example 6 : Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
- Say, Savita’s birthday can be any day of the year.Now, Hamida’s birthday can also be any day of 365 days in the year.
- Here, all
outcomes are equally likely. - (i) If Hamida’s birthday is different from Savita’s, the number of favourable outcomes is =
- So, P (Hamida’s birthday is different from Savita’s birthday) =
- P(Savita and Hamida have the same birthday) =
- We have found the answers
Moving onto next example.
Example 7 : There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?
- The number of all possible outcomes is
- The number of outcomes favourable for a card with the name of a girl =
- P(name of a girl) = P(Girl) =
25 40 - The number of outcomes favourable for a card with the name of a boy =
- P(card with name of a boy) = P(Boy) =
15 40 - We have found the answers
Note : We can also determine P(Boy), by taking P(Boy) = 1 – P(not Boy) = 1 – P(Girl) = 1-
Example 8 : A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be: (i) white? (ii) blue? (iii) red?
- Marble drawn at random means all the marbles are equally likely to be drawn.
- Number of possible outcomes =
- Let W denote the event ‘the marble is white’, B denote the event ‘the marble is blue’ and R denote the event ‘marble is red’.
- The number of outcomes favourable to the event W =
So, P(W) = - Similarly, P(B) =
= - P(R) =
- We have found the answers
Note: that P(W) + P(B) + P(R) = 1.
Example 9 : Harpreet tosses two different coins simultaneously (say, one is of Rs. 1 and other of Rs. 2). What is the probability that she gets at least one head?
- We write H for ‘head’ and T for ‘tail’. When two coins are tossed simultaneously,
- Possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely.
- Thus, the number of outcomes favourable to E is
. Therefore, P(E) = - The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) and (T, H).
- The probability that Harpreet gets at least one head is
3 4
Note : You can also find P(E) as follows: P (E) = 1 – P(
Since P(
Did you observe that in all the examples discussed so far, the number of possible outcomes in each experiment was finite?
There are many experiments in which the outcome is any number between two given numbers, or in which the outcome is every point within a circle or rectangle, etc. Can you now count the number of all possible outcomes? As you know, this is not possible since there are infinitely many numbers between two given numbers, or there are infinitely many points within a circle.
So, the definition of (theoretical) probability which you have learnt so far cannot be applied in the present form. What is the way out? To answer this, let us consider the following example :
Example 10* : In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?
- Let E be the event that ‘the music is stopped within the first half-minute’.
- Since all the outcomes are equally likely, the time period favourable to the event E is
- So, P(E) =
Time period favourable to the event E Time period in which outcomes can lie - P(E) =
1 2 2 - We have found the answer.
Now, let's solve another example involving probability of finding a crashed site.
Example 11* : A missing helicopter is reported to have crashed somewhere in the rectangular region with length 9 km and breadth 4.5 km. On the north-eastern corner, a rectangular lake with length 3 km and breadth 2.5 km is present. What is the probability that it crashed inside the lake?
- We know that the area of the rectangular region/ lake is:
keys="+ – × π ÷ brackets" where l - length and b - breadth of the considered region. - Using the formula we get: Area of the entire region where the helicopter can crash =
km 2 - Area of the lake =
km 2 - So, P (helicopter crashed in the lake) =
- We have found the answer.
Now, let's solve a problem for a shirts trader.
Example 12 : A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that:
(i) it is acceptable to Jimmy? (ii) it is acceptable to Sujatha?
- One shirt is drawn at random from the carton of 100 shirts. Therefore, there are
equally likely outcomes. - The number of outcomes favourable to Jimmy =
. Therefore, P (shirt is acceptable to Jimmy) = = - The number of outcomes favourable to Sujatha = (Number of good shirts)
+ (Number of shirts with minor defects) = - So, P (shirt is acceptable to Sujatha) =
= - We have found the answer.
Before proceeding to the next example, let's see what happens when two dice are rolled at a time.
Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on top of the dice is 8.
- E=sum of two numbers on top of dice is 8
This table shows all possible outcomes with X marking the favourable outcomes:
| X | 1 | 2 | 3 | 4 | 5 | 6 |
1 | ||||||
2 | X | |||||
3 | X | |||||
4 | X | |||||
5 | X | |||||
6 | X |
- Number of outcomes favourable to E =
- Number of all possible outcomes of the experiment=
So P(E)=
Now, let's solve a problem regarding rolling two dice.
Example 13 : Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8? (ii) 13? (iii) less than or equal to 12?
Note: The pair (1, 4) is different from (4, 1).
- The number of possible outcomes = First die outcomes × Second die outcomes = 6 × 6 =
. - Let event 'E' is favourable to the event ‘the sum of the two numbers is 8’. Thus, P(E) =
- The outcomes favourable to the event ‘the sum of the two numbers is 8’ denoted by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) i.e. the number of outcomes favourable to E = 5.
- Similarly, P(F) =
0 36 where the event F, ‘the sum of two numbers is 13’. - There is no outcome favourable to the event F.
- Let event G denoted outcomes favourable to ‘sum of two numbers is 12’. Thus, P(G) =
= - All the outcomes are favourable to the event G, ‘sum of two numbers is 12’.
- We have found the answer.
Let's look at some additional information regarding probability of multiple throws for a die.
Some additional Information
Predicting the Future
If we roll a die, the result is a number between 1 and 6, and all outcomes are equally likely. If we roll two dice at once and add up their scores we can get results from
Some results can only happen one way (to get 12 you have to roll + ) while others can happen in multiple different ways (to get 5 you could roll + or + ).
This table shows all possible outcomes:
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
The most likely result when rolling two dice is 7. There are
The least likely outcomes are 2 and 12, each with a probability of
It is impossible to forecast the outcome of a single coin toss or die roll. However, using probability we can very accurately predict the outcome of many dice.
If we throw a die 30 times, we know that we would get around
In this animation you can roll many “virtual” dice at once and see how the results compare to the predicted probabilities:
Rolling Dice
We roll
Notice how, as we roll more and more dice, the observed frequencies become closer and closer to the frequencies we predicted using probability theory. This principle applies to all probability experiments and is called the law of large numbers.
Similarly, as we increase the number of dice rolled at once, you can also see that the probabilities change from a straight line (one die) to a triangle (two dice) and then to a “bell-shaped” curve. This is known as the central limit theorem, and the bell-shaped curve is called the normal distribution.