Probability

So students, welcome to this chapter on Probability. I am really excited to teach you this chapter because probability is something we use in our daily lives without even realizing it. When we say "it might rain today" or "there is a high chance of passing the exam", we are actually talking about probability. So let's begin our journey into the world of chances and likelihoods.

In Class IX, you have already studied about experimental probability, also called empirical probability. Do you remember how we defined it? We defined it as the number of trials in which the event happened divided by the total number of trials. For example, if we toss a coin 100 times and get heads 48 times, then the experimental probability of getting a head would be 48 divided by 100, which is 0.48. This is based on what actually happens in real experiments.

Now, this experimental approach has a problem. Can you think what that might be? The problem is that we need to repeat the experiment many times to get a reliable probability. In some cases, repeating the experiment is very expensive or even impossible. For example, if we want to find the probability of a satellite failing during launch, we cannot keep launching satellites just to calculate this probability. Similarly, we cannot repeat earthquakes to find the probability of buildings getting destroyed. So we need a different approach, a theoretical approach, where we can calculate probability without actually performing the experiment many times.

This is where the theoretical probability comes into picture. But to use theoretical probability, we need to make an important assumption. We assume that all outcomes of an experiment are equally likely. Let me explain what this means.

Consider the example of a fair coin being tossed. When we say a coin is fair or unbiased, we mean that there is no reason for it to land more often on one side than the other. So when we toss such a coin, getting a head and getting a tail are equally likely outcomes. Similarly, when we throw a fair die, the outcomes 1, 2, 3, 4, 5, and 6 are all equally likely because the die is symmetrical.

But students, let me ask you something. Are the outcomes of every experiment equally likely? Suppose we have a bag containing 4 red balls and 1 blue ball, and we draw one ball without looking. Are the outcomes of drawing a red ball and drawing a blue ball equally likely? Obviously not. Since there are 4 red balls and only 1 blue ball, you are much more likely to get a red ball than a blue ball. So in this case, the outcomes are not equally likely. However, from now on in this chapter, we will assume that all experiments have equally likely outcomes. This assumption helps us calculate theoretical probability directly.

Now, let me give you the formal definition of theoretical probability. The theoretical probability, also called classical probability, of an event E, written as P(E), is defined as the number of outcomes favourable to E divided by the number of all possible outcomes of the experiment. We assume that the outcomes of the experiment are equally likely.

This definition was given by Pierre Simon Laplace in 1795. Laplace was a famous French mathematician. Interestingly, the theory of probability had its origin in the 16th century when an Italian physician and mathematician named J. Cardan wrote the first book on the subject, called "The Book on Games of Chance". Since then, many great mathematicians have contributed to this field, including James Bernoulli who lived from 1654 to 1705, A. de Moivre who lived from 1667 to 1754, and of course Laplace. Laplace's book "Theorie Analytique des Probabilités" published in 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 fields like biology, economics, genetics, physics, and sociology. So you can see how important this chapter is.

Now let's look at some examples to understand these concepts better.

Example 1: Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.

In this experiment of tossing a coin once, the number of possible outcomes is two — Head and Tail. Let E be the event 'getting a head'. The number of outcomes favourable to E, that is, of getting a head, is 1. Therefore, P(E) = P(head) = Number of outcomes favourable to E divided by Number of all possible outcomes = 1/2.

Similarly, if F is the event 'getting a tail', then P(F) = P(tail) = 1/2. This is because out of the two equally likely outcomes, one is favourable to getting a tail.

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 yellow ball? What is the probability of taking out the red ball? What is the probability of taking out the blue ball?

Since Kritika takes out a ball without looking, it is equally likely that she takes out any one of the three balls. 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'. The number of possible outcomes is 3.

For the yellow ball, the number of outcomes favourable to the event Y is 1. So P(Y) = 1/3.

Similarly, P(R) = 1/3 and P(B) = 1/3.

Now students, let me introduce you to some important terms. 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. Similarly, in Example 2, all the three events Y, B and R are elementary events.

Notice something interesting? In Example 1, P(E) + P(F) = 1/2 + 1/2 = 1. In Example 2, P(Y) + P(R) + P(B) = 1/3 + 1/3 + 1/3 = 1. So the sum of the probabilities of all the elementary events of an experiment is 1. This is true in general.

Now let's look at Example 3: Suppose we throw a die once. What is the probability of getting a number greater than 4? What is the probability of getting a number less than or equal to 4?

For the first part, let E be the event 'getting a number greater than 4'. The number of possible outcomes is six: 1, 2, 3, 4, 5 and 6. The outcomes favourable to E are 5 and 6. So the number of outcomes favourable to E is 2. Therefore, P(E) = P(number greater than 4) = 2/6 = 1/3.

For the second part, let F be the event 'getting a number less than or equal to 4'. The number of possible outcomes is still 6. The outcomes favourable to F are 1, 2, 3, 4. So the number of outcomes favourable to F is 4. Therefore, P(F) = 4/6 = 2/3.

Now students, are the events E and F in this example elementary events? No, they are not, because event E has 2 outcomes and event F has 4 outcomes. An elementary event has only one outcome.

Now let's think about something interesting. In Example 1, we had P(E) + P(F) = 1/2 + 1/2 = 1, where E is the event 'getting a head' and F is the event 'getting a tail'. In Example 3, we have P(E) + P(F) = 1/3 + 2/3 = 1, where E is the event 'getting a number greater than 4' and F is the event 'getting a number less than or equal to 4'.

Notice that getting a number not greater than 4 is the same as getting a number less than or equal to 4. So in both cases, F is actually the event 'not E'. We denote the event 'not E' by E bar. So we can write P(E) + P(not E) = 1, which gives us P(not E) = 1 – P(E).

In general, for any event E, P(E bar) = 1 – P(E). The event E bar, representing 'not E', is called the complement of the event E. We also say that E and E bar are complementary events. This is a very important result, so please remember it.

Now let me ask you two questions. First, what is the probability of getting a number 8 in a single throw of a die? Second, what is the probability of getting a number less than 7 in a single throw of a die?

For the first question, we know that there are only six possible outcomes in a single throw of a die: 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, there is no outcome favourable to getting 8. So the number of such outcomes is zero. In other words, getting 8 in a single throw of a die is impossible. So P(getting 8) = 0/6 = 0.

Students, the probability of an event which is impossible to occur is 0. Such an event is called an impossible event.

For the second question, since every face of a die is marked with a number less than 7, it is certain 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(getting a number less than 7) = 6/6 = 1.

Students, the probability of an event which is sure or certain to occur is 1. Such an event is called a sure event or a certain event.

Now note that from the definition of probability P(E), we see that the numerator, that is the number of outcomes favourable to the event E, is always less than or equal to the denominator, that is the number of all possible outcomes. Therefore, 0 ≤ P(E) ≤ 1. This is a very important property of probability. The probability of any event always lies between 0 and 1, inclusive.

Now let's look at some more examples involving playing cards. Have you seen a deck of playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards each: spades, hearts, diamonds and clubs. Clubs and spades are of black colour, while hearts and diamonds are of red colour. 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 be an ace, and also calculate the probability that the card will not be an ace.

Well-shuffling ensures equally likely outcomes. There are 4 aces in a deck. Let E be the event 'the card is an ace'. The number of outcomes favourable to E is 4. The number of possible outcomes is 52. Therefore, P(E) = 4/52 = 1/13.

Now let F be the event 'card drawn is not an ace'. The number of outcomes favourable to F is 52 minus 4, which is 48. The number of possible outcomes is 52. Therefore, P(F) = 48/52 = 12/13.

Notice that F is nothing but E bar. So we can also calculate P(F) as follows: P(F) = P(E bar) = 1 – P(E) = 1 – 1/13 = 12/13. This is exactly what we got. So using the complement rule makes calculations easier sometimes.

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 is P(S) = 0.62, given. Now, either Sangeeta wins or Reshma wins, so these are complementary events. Therefore, the probability of Reshma's winning is P(R) = 1 – P(S) = 1 – 0.62 = 0.38.

Example 6: Savita and Hamida are friends. What is the probability that both will have different birthdays? What is the probability that they will have the same birthday? We are ignoring leap years here.

Out of the two friends, one girl, 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. We assume that these 365 outcomes are equally likely.

For different birthdays, if Hamida's birthday is different from Savita's, the number of favourable outcomes for her birthday is 365 minus 1, which is 364. So P(their birthdays are different) = 364/365.

For the same birthday, P(they have the same birthday) = 1 – P(their birthdays are different) = 1 – 364/365 = 1/365.

This is a surprising result, isn't it? Even with just two people, the probability of having the same birthday is quite low, only 1 out of 365.

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 a girl? What is the probability that it is the name of a boy?

There are 40 students, and only one name card has to be chosen. So the number of all possible outcomes is 40.

For a girl, the number of outcomes favourable is 25. Therefore, P(card with name of a girl) = 25/40 = 5/8.

For a boy, the number of outcomes favourable is 15. Therefore, P(card with name of a boy) = 15/40 = 3/8.

Note that we can also determine P(boy) by taking P(boy) = 1 – P(not boy) = 1 – P(girl) = 1 – 5/8 = 3/8.

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 white? What is the probability that it will be blue? What is the probability that it will be red?

Saying that a marble is drawn at random is a short way of saying that all the marbles are equally likely to be drawn. Therefore, the number of possible outcomes is 3 + 2 + 4 = 9.

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'.

For white marble, the number of outcomes favourable to W is 2. So P(W) = 2/9.

For blue marble, P(B) = 3/9 = 1/3.

For red marble, P(R) = 4/9.

Notice that P(W) + P(B) + P(R) = 2/9 + 3/9 + 4/9 = 9/9 = 1. This again confirms that the sum of probabilities of all elementary events is 1.

Example 9: Harpreet tosses two different coins simultaneously, say one is of ₹1 and the other of ₹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, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely. Here (H, H) means head up on the first coin, that is the ₹1 coin, and head up on the second coin, that is the ₹2 coin. Similarly (H, T) means head up on the first coin and tail up on the second coin, and so on.

The outcomes favourable to the event E, 'at least one head', are (H, H), (H, T) and (T, H). These are three outcomes. So the number of outcomes favourable to E is 3.

Therefore, P(E) = 3/4. So the probability that Harpreet gets at least one head is 3/4.

You can also find P(E) as follows: P(at least one head) = 1 – P(no head) = 1 – P(getting two tails) = 1 – 1/4 = 3/4. This is using the complement rule again.

Now students, did you observe that in all the examples we have discussed so far, the number of possible outcomes in each experiment was finite? Let me ask you something. Are there experiments where the outcome can be any number between two given numbers, or where the outcome can be any point within a circle or rectangle? Can you count the number of all possible outcomes in such cases? This is not possible since there are infinitely many numbers between two given numbers, or infinitely many points within a circle. So the definition of theoretical probability that we have learnt so far cannot be applied in the present form. What is the way out? Let's see.

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?

Here the possible outcomes are all the numbers between 0 and 2. This is the portion of the number line from 0 to 2. Let E be the event that 'the music is stopped within the first half-minute'. The outcomes favourable to E are points on the number line from 0 to 1/2. The distance from 0 to 2 is 2, while the distance from 0 to 1/2 is 1/2. Since all the outcomes are equally likely, we can argue that of the total distance of 2, the distance favourable to the event E is 1/2. So P(E) = (1/2)/2 = 1/4.

So students, we can extend this idea of probability as the ratio of favourable length to total length, or more generally, the ratio of favourable area to total area.

Example 11: A missing helicopter is reported to have crashed somewhere in a rectangular region. What is the probability that it crashed inside the lake shown in the figure?

The helicopter is equally likely to crash anywhere in the region. The area of the entire region where the helicopter can crash is (4.5 × 9) square kilometres, which equals 40.5 square kilometres. The area of the lake is (2.5 × 3) square kilometres, which equals 7.5 square kilometres. Therefore, P(helicopter crashed in the lake) = 7.5/40.5 = 75/405 = 5/27.

So you see, even when there are infinitely many possible outcomes, we can still calculate probability by using areas.

Now let's look at some more examples.

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 it is acceptable to Jimmy? What is the probability that it is acceptable to Sujatha?

One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes.

For Jimmy, the number of outcomes favourable, that is acceptable shirts, is 88. Therefore, P(shirt is acceptable to Jimmy) = 88/100 = 0.88.

For Sujatha, she will accept any shirt that does not have major defects. So she will accept both good shirts and shirts with minor defects. That is 88 + 8 = 96 shirts. So P(shirt is acceptable to Sujatha) = 96/100 = 0.96.

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 8? What is the probability that the sum is 13? What is the probability that the sum is less than or equal to 12?

When the blue die shows '1', the grey die could show any one of the numbers 1, 2, 3, 4, 5, 6. The same is true when the blue die shows '2', '3', '4', '5' or '6'. The possible outcomes of the experiment are listed in a table; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die.

Note that the pair (1, 4) is different from (4, 1) because the blue die and the grey die are distinguishable. So the number of possible outcomes is 6 multiplied by 6, which equals 36.

For the sum to be 8, the favourable outcomes are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2). So the number of outcomes favourable is 5. Hence, P(sum is 8) = 5/36.

For the sum to be 13, there is no outcome favourable because the maximum sum possible with two dice is 6 + 6 = 12. So P(sum is 13) = 0/36 = 0. This is an impossible event.

For the sum to be less than or equal to 12, all outcomes are favourable because the minimum sum is 1 + 1 = 2 and the maximum sum is 12. So P(sum ≤ 12) = 36/36 = 1. This is a sure event.

Now students, let me summarize what we have learned in this chapter.

First, we learned about equally likely outcomes. When we perform an experiment, we say outcomes are equally likely if each outcome has the same chance of occurring. For example, when we toss a fair coin, getting head and getting tail are equally likely. When we throw a fair die, all six faces are equally likely to come up.

Second, we learned the definition of theoretical probability. The probability of an event E, written as P(E), is defined as the number of outcomes favourable to E divided by the number of all possible outcomes of the experiment, assuming that the outcomes are equally likely.

Third, we learned about elementary events. An elementary event is an event that has only one outcome. For example, when we toss a coin, getting a head is an elementary event.

Fourth, we learned about complementary events. For any event E, the event 'not E' is called the complement of E, denoted by E bar. The important relation is P(E) + P(E bar) = 1, or P(E bar) = 1 – P(E).

Fifth, we learned about impossible events and sure events. The probability of an impossible event is 0, and the probability of a sure event is 1.

Sixth, we learned that for any event E, 0 ≤ P(E) ≤ 1. Probability always lies between 0 and 1, inclusive.

Seventh, we learned that the sum of probabilities of all elementary events of an experiment is always 1.

We also learned how to apply these concepts to various situations: tossing coins, throwing dice, drawing cards from a deck, drawing balls from bags, and even situations involving geometry like finding probabilities based on areas.

We also saw that probability theory has a rich history, starting from the 16th century with J. Cardan, and developed further by mathematicians like Bernoulli, de Moivre, and Laplace. Today, probability is used in many fields including biology, economics, genetics, physics, and sociology.

One important thing to remember is the difference between experimental probability and theoretical probability. Experimental probability is based on what has actually happened, while theoretical probability attempts to predict what will happen based on certain assumptions. As the number of trials in an experiment increases, we may expect the experimental and theoretical probabilities to be nearly the same.

Students, this concludes our lesson on Probability. You have learned about the theoretical approach to probability, how to calculate probability when outcomes are equally likely, and how to apply these concepts to solve various problems. Make sure you understand each concept clearly and practice as many problems as you can. Thank you for your attention, and best of luck with your studies!