**Introduction**

Galileo, an Italian mathematician was the first person to attempt at a quantitative measure of probability while dealing with some problems related to the theory of dice in gambling. In 1654, a French gambler and nobleman Chevalier De-Mere, approached French philosopher and mathematician Blaise Pascal regarding some dice problems. The famous “ problem of points “ posed by De-Mere to Pascal is “ Two persons play a game of chance. The person who first gains a certain number of points wins the stake. They stop playing before the game is completed. How is the stake to be decided on the basis of the number of points each has won? “ The discussion of this problem between Blaise Pascal and another French mathematician Perre de Fermat laid the foundation of probability. Let us learn more about probability and various instances of finding probability on rolling a die.

**Definition**

Let n be the total number of trials of a random experiment. Then the probability P ( E ) of happening of an event E is defined as –

P ( E ) = $\frac{Number\: of\: trials\: in\: which\: the\: event\: happened}{Total\: number\: of\: trials}$

Let us recall some basic definitions related to probability.

**Trial **– A trial is an action that results in one or several outcomes.

**Random Experiment** – An experiment in which the result of a trial cannot be predicted in advance is called a random experiment.

**Event** – An event associated to a random experiment is the collection of some outcomes of the experiment.

**Occurrence or Happening of an event** – An event associated with a random experiment is said to occur or happen if any one of the outcomes satisfying the definition of the event is an outcome of the experiment when it is performed.

**Compound Event** – A collection of two or more possible outcomes of a trial of a random experiment I called a compound event.

**What is meant by rolling a die?**

In a random experiment of throwing a die, if the outcome of the throw is 4, then we say that each one of the following happens –

- Getting 4
- Getting an even number
- Getting a number greater than 3

However, on the basis of the same outcome, one can say that none of the following events have happened –

- Getting an odd number
- Getting a multiple of 3
- Getting a number less than 3
- Getting a prime number

On the basis of these discussions, we define the experimental or empirical probability of happening of an event.

Let us consider an example.

**Example**

A die is thrown 1000 times with the following frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given below –

Outcome : | 1 | 2 | 3 | 4 | 5 | 6 |

Frequency : | 179 | 150 | 157 | 149 | 175 | 190 |

Find the probability of happening of each outcome.

**Solution**

We have been given that a die is thrown 1000 times with the following frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given below –

Outcome : | 1 | 2 | 3 | 4 | 5 | 6 |

Frequency : | 179 | 150 | 157 | 149 | 175 | 190 |

We need to find the probability of happening of each outcome.

Let A _{i} denote the event of getting the outcome of i, where i = 1, 2, 3, 4, 5, 6. Then,

P ( E _{1} ) = Probability of getting outcome 1 = $\frac{Number\: of\: trials\: in\: which\: the\: event\: happened}{Total\: number\: of\: trials}$

⇒ P ( E _{1} ) = $\frac{Frequency\: of\: 1}{Total\: number\: of\: times\: the\: die\: is\: thrown}$

⇒ P ( E _{1} ) = $\frac{1 7 9}{1 0 0 0}$ = 0. 1 7 9 . . . . . . . . . . . . . . . . . . . ( 1 )

P ( E _{2} ) = Probability of getting outcome 2 = $\frac{Number\: of\: trials\: in\: which\: the\: event\: happened}{Total\: number\: of\: trials}$

⇒ P ( E _{2} ) = $\frac{Frequency\: of\: 2}{Total\: number\: of\: times\: the\: die\: is\: thrown}$

⇒ P ( E _{2} ) = $\frac{1 5 0}{1 0 0 0}$ = 0. 1 5 0 . . . . . . . . . . . . . . . . . . . ( 2 )

P ( E _{3} ) = Probability of getting outcome 3 = $\frac{Number\: of\: trials\: in\: which\: the\: event\: happened}{Total\: number\: of\: trials}$

⇒ P ( E _{3} ) = $\frac{Frequency\: of\: 3}{Total\: number\: of\: times\: the\: die\: is\: thrown}$

⇒ P ( E _{3} ) = $\frac{1 5 7}{1 0 0 0}$ = 0. 1 5 7 . . . . . . . . . . . . . . . . . . . ( 3 )

P ( E _{4} ) = Probability of getting outcome 4 = $\frac{Number\: of\: trials\: in\: which\: the\: event\: happened}{Total\: number\: of\: trials}$

⇒ P ( E _{4} ) = $\frac{Frequency\: of\: 4}{Total\: number\: of\: times\: the\: die\: is\: thrown}$

⇒ P ( E _{4} ) = $\frac{1 4 9}{1 0 0 0}$ = 0. 1 4 9 . . . . . . . . . . . . . . . . . . . ( 4 )

P ( E _{5} ) = Probability of getting outcome 5 = $\frac{Number\: of\: trials\: in\: which\: the\: event\: happened}{Total\: number\: of\: trials}$

⇒ P ( E _{5} ) = $\frac{Frequency\: of\: 5}{Total\: number\: of\: times\: the\: die\: is\: thrown}$

⇒ P ( E _{5} ) = $\frac{1 7 5}{1 0 0 0}$ = 0. 1 7 5 . . . . . . . . . . . . . . . . . . . ( 5 )

P ( E _{6} ) = Probability of getting outcome 6 = $\frac{Number\: of\: trials\: in\: which\: the\: event\: happened}{Total\: number\: of\: trials}$

⇒ P ( E _{6} ) = $\frac{Frequency\: of\: 6}{Total\: number\: of\: times\: the\: die\: is\: thrown}$

⇒ P ( E _{6} ) = $\frac{1 9 0}{1 0 0 0}$ = 0. 1 9 0 . . . . . . . . . . . . . . . . . . . ( 6 )

From ( 1 ), ( 2 ) , ( 3 ) , ( 4 ), ( 5 ) and ( 6 ), we have,

**P ( E **_{1}** ) = 0. 1 7 9 , P ( E **_{2}** ) = 0. 1 5 0 , P ( E **_{3}** ) = 0. 1 5 7 , P ( E **_{4}** ) = 0. 1 4 9 , P ( E **_{5}** ) = 0. 1 7 5 and P ( E **_{6}** ) = 0. 1 9 0 **

**Sample Space**

The set of all possible outcomes of a random experiment is called the sample space associated with it and is generally denoted by S. This means that if E _{1}, E _{2}, E_{3}, . . . . . . . . . .E _{n} are the possible outcomes of a random experiment then S = { E _{1}, E _{2}, E_{3}, . . . . . . . . . .E _{n }} is the sample space associated to it.

Let us understand it through an example.

Consider a random experiment where two die are tossed together or a die is tossed twice. If we define,

E _{ij} = Getting number i on the upper face of the first die and the number j on the upper face of the second die, where i = 1, 2, 3, 4, 5, 6 and j = 1, 2, 3, 4, 5, 6

Then E _{ij} are elementary events associated to this experiment ad are generally denoted by ( i, j )

Thus ( 1, 1 ), ( 1, 2 ), ( 1 , 3 ) , ( 1 , 4 ) , ( 1, 5 ) , ( 1 , 6 ) , ( 2, 1 ), ( 2, 2 ), ( 2 , 3 ) , ( 2 , 4 ) , ( 2, 5 ) , ( 2 , 6 ), ( 3, 1 ), ( 3, 2 ), ( 3 , 3 ) , ( 3 , 4 ) , ( 3, 5 ) , ( 3 , 6 ), ( 4, 1 ), ( 4, 2 ), ( 4 , 3 ) , ( 4 , 4 ) , ( 4, 5 ) , ( 4 , 6 ), ( 5, 1 ), ( 5, 2 ), ( 5 , 3 ) , ( 5 , 4 ) , ( 5, 5 ) , ( 5 , 6 ), ( 6, 1 ), ( 6, 2 ), ( 6 , 3 ) , ( 6 , 4 ) , ( 6, 5 ) , ( 6 , 6 ) are 36 elementary events associated to the random experiment of tossing two die and the sample space associated to it is given by

S = { ( 1, 1 ), ( 1, 2 ), ( 1 , 3 ) , ( 1 , 4 ) , ( 1, 5 ) , ( 1 , 6 ) , ( 2, 1 ), ( 2, 2 ), ( 2 , 3 ) , ( 2 , 4 ) , ( 2, 5 ) , ( 2 , 6 ), ( 3, 1 ), ( 3, 2 ), ( 3 , 3 ) , ( 3 , 4 ) , ( 3, 5 ) , ( 3 , 6 ), ( 4, 1 ), ( 4, 2 ), ( 4 , 3 ) , ( 4 , 4 ) , ( 4, 5 ) , ( 4 , 6 ), ( 5, 1 ), ( 5, 2 ), ( 5 , 3 ) , ( 5 , 4 ) , ( 5, 5 ) , ( 5 , 6 ), ( 6, 1 ), ( 6, 2 ), ( 6 , 3 ) , ( 6 , 4 ) , ( 6, 5 ) , ( 6 , 6 ) }

**Solved Examples**

**Example 1** An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.

**Solution** If the die is rolled and we get an even number ( 2 or 4 or 6 ) on its upper face, then we toss a coin which may result in head ( H ) or tail ( T ). So the possible outcomes in this case are –

(2, H ) , ( 4, H ) , ( 6, H ) , (2, T ) , ( 4, T ) , ( 6, T ) . . . . . . . . . ( 1 )

If the die is rolled and we get and odd number ( 1 or 3 or 5 ) on its upper face then the coin is tossed twice which may result in one of the following ways: HH, HT, TH, TT. So, the possible outcomes in this case are –

( 1, HH ), ( 3, HH ), ( 5, HH ), ( 1, HT ), ( 3, HT ) ( 5, HT ), ( 1, TH ), ( 3, TH ) ( 5, TH ), ( 1, TT ), ( 3, TT ) ( 5, TT ) . . . . . . . . . . . . ( 2 )

From ( 1 ) and ( 2 ), we can say that all elementary events associated to the experiment are –

(2, H ) , ( 4, H ) , ( 6, H ) , (2, T ) , ( 4, T ) , ( 6, T ), ( 1, HH ), ( 3, HH ), ( 5, HH ), ( 1, HT ), ( 3, HT ) ( 5, HT ), ( 1, TH ), ( 3, TH ) ( 5, TH ), ( 1, TT ), ( 3, TT ) ( 5, TT )

Therefore, the sample space associated with the random experiment is

S = { (2, H ) , ( 4, H ) , ( 6, H ) , (2, T ) , ( 4, T ) , ( 6, T ), ( 1, HH ), ( 3, HH ), ( 5, HH ), ( 1, HT ), ( 3, HT ) ( 5, HT ), ( 1, TH ), ( 3, TH ) ( 5, TH ), ( 1, TT ), ( 3, TT ) ( 5, TT ) }

**Example 2** An unbiased die is thrown. What is the probability of getting

a) an even number

b) a multiple of 3

c) a number less than 5

**Solution** We have been given that an unbiased die is thrown. In a single throw of die we can get any one of the six numbers, 1, 2, 3, 4, 5, 6 marked on its six faces. Therefore, the total number of elementary events associated with the random experiment of throwing a die is 6.

a) Let A denote the probability of getting an even number. Now, event A will be said to have occurred if we get any one of 2, 4 or 6 as an outcome.

Therefore, Favourable number of events = 3

Now, we know that

Probability = $\frac{Favourable\: Number\: of\” outcomes}{Total\: number\: of\: outcomes}$

Hence, the probability of getting an even number P ( A ) = $\frac{3}{6} = \frac{1}{2}$

b) Let A denote the probability of getting a multiple of 3. Now, event A will be said to have occurred if we get any one of 3 or 6 as an outcome.

Therefore, Favourable number of events = 2

Now, we know that

Probability = $\frac{Favourable\: Number\: of\” outcomes}{Total\: number\: of\: outcomes}$

Hence, the probability of getting a multiple of 3, P ( A ) = $\frac{2}{6} = \frac{1}{3}$

c) Let A denote the probability of getting a number less than 5. Now, event A will be said to have occurred if we get any one of 1, 2, 3, 4 as an outcome.

Therefore, Favourable number of events = 4

Now, we know that

Probability = $\frac{Favourable\: Number\: of\” outcomes}{Total\: number\: of\: outcomes}$

Hence, the probability of getting a number less than 5, P ( A ) = $\frac{4}{6} = \frac{2}{3}$

**Key Facts and Summary**

- Let n be the total number of trials of a random experiment. Then the probability

P ( E ) of happening of an event E is defined as – P ( E ) = $\frac{Number\: of\: trials\: in\: which\: the\: event\: happened}{Total\: number\: of\: trials}$ - A trial is an action that results in one or several outcomes.
- An experiment in which the result of a trial cannot be predicted in advance is called a random experiment.
- An event associated to a random experiment is the collection of some outcomes of the experiment.
- An event associated with a random experiment is said to occur or happen if any one of the outcomes satisfying the definition of the event is an outcome of the experiment when it is performed.
- A collection of two or more possible outcomes of a trial of a random experiment I called a compound event.
- The set of all possible outcomes of a random experiment is called the sample space associated with it and is generally denoted by S. This means that if E
_{1}, E_{2}, E_{3}, . . . . . . . . . .E_{n}are the possible outcomes of a random experiment then S = { E_{1}, E_{2}, E_{3}, . . . . . . . . . .E_{n }} is the sample space associated to it.

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