Random experiments and sample spaces

What you will be able to do

• say what a random experiment, an outcome and a sample space are;
• write and count the sample space of coins, dice and cards;
• count the outcomes of drawing from a bag with replacement, without replacement and together.

The toss before the game

Before a game, a referee tosses a coin. Nobody can say whether it will show a head or a tail. But everybody can say what it might show: a head or a tail, and nothing else.

That is the whole idea of this lesson. We cannot predict one result. We can list all of them.

Words we need

• A random experiment is an action that can be repeated under the same conditions, where each time one result from a known list occurs, and the result cannot be told in advance.
• Each possible result is an outcome.
• The set of all outcomes is the sample space, written $S$. The number of outcomes is $n(S)$.

For one toss of a coin, $S = \{H, T\}$ and $n(S) = 2$. For one throw of a die, $S = \{1, 2, 3, 4, 5, 6\}$ and $n(S) = 6$.

Most of the time in this chapter we use equally likely outcomes. That means a fair coin, a fair die, a well-shuffled pack of cards, and balls that are alike in size and weight. Then no outcome is favoured over another.

More than one thing at a time

A coin tossed three times is one experiment with a longer list. We write each outcome as a string, in the order of the tosses:

So $n(S) = 8$. Each toss has $2$ results, and the results multiply: $2 \times 2 \times 2 = 8$. This is the multiplication principle from the chapter on permutations and combinations.

For two dice, an outcome is a pair $(x, y)$. The first number is the first die, and the second number is the second die. There are $6 \times 6 = 36$ pairs, and $(2, 5)$ is a different outcome from $(5, 2)$.

A trap. Suppose the two dice look exactly alike. A student may say that only $21$ pairs are possible, because $(2, 5)$ and $(5, 2)$ look the same. The $21$ pairs are real, but they are not equally likely. The pair $(2, 5)$ can happen in two ways and the pair $(3, 3)$ in one way. To keep every outcome equally likely, always think of the dice as the first die and the second die, and use the $36$ ordered pairs.

Cards

An ordinary pack has $52$ cards. There are four suits: spades and clubs are black, hearts and diamonds are red. Each suit has $13$ cards: ace, $2$ to $10$, jack, queen and king. The jack, queen and king are the face cards, so a pack holds $12$ face cards. Drawing one card gives a sample space of $52$ outcomes.

A bag of balls

A bag holds a red ball $R$, a green ball $G$ and a blue ball $B$. Two balls are taken out. How many outcomes are there? The answer depends on how you take them out.

1. With replacement. Take a ball, note it, put it back, then take again. Any ball can come both times. The outcomes are ordered pairs, and $n(S) = 3 \times 3 = 9$.
2. Without replacement, one after the other. The first ball is not put back. The outcomes are still ordered pairs, but a ball cannot repeat. So $n(S) = 3 \times 2 = 6$.
3. Two taken together. Order does not matter, because nobody sees which came first. The outcomes are pairs without order: $\{R, G\}$, $\{R, B\}$, $\{G, B\}$. So $n(S) = \dfrac{3 \times 2}{2} = 3$.

In general, drawing $r$ balls from $n$ different balls gives $n^r$ outcomes with replacement, $n(n-1)\cdots(n-r+1)$ in order without replacement, and that number divided by $r!$ when the order is ignored. Do you see $^nP_r$ and $^nC_r$ from the Class 11 chapter on permutations and combinations? They are the same counts.

A worked example

A coin is tossed and then a die is thrown. Write the sample space and count the outcomes in which the coin shows a head and the die shows a number greater than $4$.

1. The sample space. Pair each result of the coin with each result of the die: $H1, H2, \dots, H6, T1, T2, \dots, T6$. So $n(S) = 2 \times 6 = 12$.
2. The outcomes we want. The coin shows $H$ and the die shows $5$ or $6$. They are $H5$ and $H6$.
3. The count. There are $2$ such outcomes out of $12$.

Practise with new numbers

Try it in Python

The program lists the sample spaces from this lesson and counts them. The last lines show why we do not use the $21$ pairs of two identical dice.

Common mistakes

• Treating $(2, 5)$ and $(5, 2)$ as one outcome. The two dice are the first and the second die. The pairs are ordered.
• Counting "number of heads" as the outcome. For three tosses, "$2$ heads" is not one outcome. It is the three outcomes $HHT$, $HTH$ and $THH$.
• Adding where we should multiply. Three tosses give $2 \times 2 \times 2$ outcomes, not $2 + 2 + 2$.
• Using the wrong draw. "With replacement", "without replacement" and "together" give different sample spaces. Read the words of the question.

Remember these

• The sample space $S$ is the set of all outcomes, and $n(S)$ is how many there are.
• Independent steps multiply: three tosses give $2^3 = 8$ and two dice give $6^2 = 36$.
• A pack has $52$ cards: four suits of $13$, with $12$ face cards.
• A draw of $r$ from $n$: $n^r$ with replacement, $^nP_r$ in order without replacement, $^nC_r$ together.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How many outcomes does the sample space have when a coin is tossed four times?

2. A bag holds $4$ different balls. Two are taken out together. How many outcomes are there?

3. A die is thrown and then a card is drawn from a pack. How many outcomes are in the sample space?

4. A student says that two dice have $21$ outcomes. What is wrong with using these $21$ pairs?

5. List the outcomes of tossing three coins in which exactly two show heads.

References

• NCERT, Mathematics for Class XI (earlier edition), Probability, section on random experiments and the sample space.
• OpenStax, Introductory Statistics, Chapter 3, section on terminology.