Probability

Why This Matters

Before a cricket match, the two captains toss a coin. Everyone agrees this is fair. But why? A fair coin has the same chance of landing heads as tails. So neither team gets an advantage. That simple idea is the heart of this whole chapter.

Sometimes a few results are equally likely. “Equally likely” just means each result has the same chance of happening, like heads and tails on a fair coin. When results are equally likely, we can give an exact number to “how likely” each one is.

That number is called a probability. It tells you the chance of something happening. It always sits between 0 and 1. A probability of 0 means it never happens. A probability of 1 means it always happens. Probability is everywhere. Weather forecasts use it. Insurance companies use it to decide prices. Games use it for odds. Even scientists use it to guess if a satellite might fail. Doctors and economists use it too. Whenever the future is uncertain, probability helps people decide.

You actually met probability once before, in Class 9, but you found it a very different way. Let’s quickly refresh how that worked before we build on it.

In Class 9 you found probability by doing an experiment many times and counting. For example, you might toss a coin 1000 times and count how often heads came up. That works, but it has a problem. You cannot always repeat an experiment. You cannot launch a satellite a thousand times just to see how often it fails. So this chapter shows a smarter way. When the results are equally likely, you can calculate the probability directly. You do not need to do any experiment at all.

The Big Idea

Suppose an experiment has a fixed list of equally likely outcomes. Then the theoretical (classical) probability of an event E is just a count. You divide two numbers: P(E) = (number of outcomes that give you E) / (total number of possible outcomes). Every probability sits between 0 and 1. An impossible event has probability 0. A sure event has probability 1. And the chance of E not happening makes up the rest: P(not E) = 1 − P(E).

Let’s Break It Down

What theoretical probability is

First, let’s learn three words we will use a lot.

An experiment is an action that can have more than one result. Tossing a coin is an experiment. So is throwing a die or drawing a card.

Each possible result is called an outcome. For a coin, the outcomes are heads and tails. For a die, the outcomes are 1, 2, 3, 4, 5 and 6.

An event is any group of outcomes that you care about. For example, “getting an even number” on a die is an event. It is made of three outcomes: 2, 4 and 6.

This whole method needs one thing to be true. The outcomes must be equally likely. That means no outcome has a better chance than another. A fair coin gives equally likely outcomes. (Another word for “fair” is unbiased — it just means the coin is not loaded to favour one side.) A fair die and a well-shuffled deck of cards also give equally likely outcomes. When the outcomes are equally likely, we can find the probability of an event E directly:

P(E) = (number of outcomes favourable to E) / (total number of all possible outcomes)

The word “favourable” just means the outcomes that give you the event you want. So the rule is simple: favourable divided by total. No experiment needed. Take one toss of a coin. The two equally likely outcomes are Head and Tail. The event “head” has 1 favourable outcome out of 2 total. So P(head) = 1/2. In the same way, P(tail) = 1/2.

But why does that little “equally likely” condition matter so much? It is the hidden hinge the whole formula swings on. Counting “favourable out of total” only tells the truth if each outcome carries the same chance. Figure 14.1 below makes this clear with two spinners — a fair one with equal slices, and a loaded one with unequal slices.

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Look at the fair spinner. All six slices are the same size, so the arrow is just as likely to stop on any number. Here “1 out of 6 slices” genuinely means a chance of 1/6 — the count and the real chance match. Now look at the loaded spinner. The slice for 1 is huge and the others are tiny. There are still six numbers, so the lazy answer would be “1/6 each”. But that is plainly false — the arrow lands on 1 far more often. The formula breaks because it only counts how many slices there are, not how big each one is. When the slices are equal, size and count say the same thing, so the formula is safe. When they are not, you cannot just count. This is exactly why we insist the coin is fair, the die is fair, and the deck is well-shuffled: those words are our promise that the outcomes are equally likely, so counting is allowed.

Notice that every answer above came out as a fraction. Since probabilities are fractions, it helps to recall how to tidy a fraction into its neatest form.

Let’s put the favourable-over-total rule to work on a real example. Here a fair die is thrown, and we find two probabilities that turn out to be linked.

An event that has only one outcome is called an elementary event. “Getting a head” is an elementary event. So is “the die shows 3”. Here is a neat fact: if you add up the probabilities of all the elementary events, you always get 1. For the die, each face has probability 1/6, and there are 6 faces. So P(1) + P(2) + … + P(6) = 1/6 × 6 = 1.

The 0-to-1 range: sure and impossible events

The number of favourable outcomes can never be less than 0. It can also never be more than the total. So every probability is squeezed between 0 and 1:

0 ≤ P(E) ≤ 1

Why can it never break out of that range? It comes straight from the formula, and Figure 14.2 below shows the idea. The favourable outcomes are always just a part of the total outcomes — they are picked from the same list, not from somewhere else.

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So the bottom number (total) is always the biggest the top number (favourable) can ever be. If no outcome is favourable, the top is 0 and P = 0/total = 0. If every outcome is favourable, the top equals the bottom and P = total/total = 1. You can never count a negative number of outcomes, and you can never count more favourable outcomes than there are outcomes in the first place. That is the whole reason every probability lands somewhere from 0 to 1, and never outside.

A good way to picture this is a line that runs from 0 to 1, as in Figure 14.3 below. Every probability is a point on this line. At the far left is 0. This is where the impossible events sit. At the far right is 1. This is where the sure events sit. Everything that might happen sits somewhere in between. An even chance lands right in the middle, at 1/2.

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The two ends of the line have special names. An impossible event can never happen. It has 0 favourable outcomes, so P = 0. For example, throwing one die and getting an 8 is impossible. No face has an 8 on it. So P(getting 8) = 0/6 = 0. A sure event (also called a certain event) always happens. Every outcome is favourable, so P = 1. For example, getting a number less than 7 on a die is certain. All six faces are less than 7. So P(less than 7) = 6/6 = 1.

Here are the three kinds of event side by side, so you can see at a glance how many outcomes are favourable in each and where its probability lands.

Try spotting an impossible and a sure event yourself with this quick candy question.

Complementary events: P(not E) = 1 − P(E)

Every event E has a partner event called “not E”. It is made of everything that is not in E. We write it as E̅. Think of it like this. If E is “getting a head”, then not E is “getting a tail”. Together, E and E̅ cover all the outcomes, and they never overlap. So their probabilities must add up to the whole:

P(E) + P(not E) = 1. If we move things around, this becomes P(not E) = 1 − P(E).

E and E̅ are called complementary events. This small formula is a big shortcut. Often the event you want is hard to count directly. But its complement is easy to count. So instead of counting the hard one, you count the easy complement and subtract it from 1.

Figure 14.4 below shows the idea as a single bar of length 1, split into the part for E and the part for not E.

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That card example only works if you know exactly what is in a deck. Since cards appear in so many problems, here is the full make-up of the pack to keep handy.

Here is an example. You draw a card from a deck and want P(not an ace). Counting all the non-ace cards one by one is slow. There are 48 of them! But the complement is easy. There are 4 aces in 52 cards, so P(ace) = 4/52 = 1/13. So P(not an ace) = 1 − 1/13 = 12/13. That was much faster.

The complement isn’t just for cards. In this match example, knowing one player’s chance instantly hands us the other’s.

Worked examples: coins, dice and cards

These three setups come up again and again. So it really helps to know each one’s list of outcomes well. A coin has 2 outcomes: Head and Tail. A die has 6 outcomes: the faces 1 to 6. A standard deck has 52 cards. The cards are split into 4 suits of 13 cards each. Spades and clubs are black. Hearts and diamonds are red. Each suit has these cards: ace, king, queen, jack, 10, 9, and so on down to 2. The kings, queens and jacks are called face cards. There are 12 face cards in all (3 in each of the 4 suits).

Let’s start with two coins at once. The trick is to write each result as a pair, and watch how “at least one head” can be counted two ways.

Now over to the cards. This next example finds three probabilities from one draw — a broad event, a medium one, and a single exact card.

Now let’s look at the famous two-dice problem. When you throw two dice, each outcome is a pair: (first die, second die). The order matters here. So 1 on the first die and 4 on the second is a different outcome from 4 on the first and 1 on the second. The first die can show 6 numbers, and for each of those, the second die can also show 6 numbers. So there are 6 × 6 = 36 equally likely outcomes. The easiest way to see them all is in a grid, shown in Figure 14.5 below:

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Using that grid of 36 outcomes, let’s find three sums — one ordinary, one impossible, and one that is certain.

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Summary

You should now be able to explain:

• The theoretical (classical) probability of an event E is P(E) = (number of favourable outcomes) / (total number of all possible outcomes). This only works if the outcomes are equally likely.
• The probability of a sure (certain) event is 1. The probability of an impossible event is 0.
• Every probability lies in the range 0 ≤ P(E) ≤ 1.
• An elementary event has just one outcome. The probabilities of all the elementary events of an experiment add up to 1.
• For any event E, P(E) + P(not E) = 1, so P(not E) = 1 − P(E). E and not-E are complementary events. The complement is often the quick way to an answer.
• Know your outcome lists by heart: a coin has 2 outcomes, a die has 6, two dice have 36 pairs, and a deck has 52 cards (4 suits of 13, with 12 face cards).
• Experimental probability (from Class 9) comes from actually doing the experiment many times. Theoretical probability predicts the chance using assumptions instead. As you do more and more trials, the two values get closer and closer.

What’s Next

This is the last chapter of your Class 10 Maths journey. So it is a good time to look back. Notice how this chapter pulls together ideas from the whole year. Every P(E) is a fraction, which you have used since the start. The careful listing and counting here is the same kind of logic you met in earlier chapters. And being comfortable with 0, 1 and every number in between goes right back to real numbers. You have built up a lot of tools this year. Real Numbers, Polynomials, Linear Equations, Quadratics, Trigonometry, Coordinate Geometry, Areas and Volumes, Statistics, and now Probability — each one is a different way to describe the world clearly. Probability is special for one reason. It lets you reason with confidence even when you cannot be sure. That is exactly the situation in most real decisions. So keep using this maths. Estimate, model, check, and predict. You now have a full toolkit. Go use it.