Let us now summarize the terminology we have discussed up to this point.

Random Experiment. A random experiment is an experiment for which the set of possible outcomes is known, but which particular outcome will occur on a particular execution of the experiment cannot be said prior to performing the experiment. Tossing a coin, rolling a die and drawing a card from a deck are all examples of random experiments.

Outcome. An outcome of a random experiment is any one of the possible results of the experiment. For example, in rolling a die, “3” is a possible outcome. In tossing two coins simultaneously, {H, T} is a possible outcome.

Sample space. The sample space of a random experiment is the set of all possible outcomes for that experiment. Some examples:

Tossing a coin: the sample space is {H, T}

Rolling a die: the sample space the six outcomes {1, 2, 3, 4, 5, 6}.

Drawing a card from a deck: the sample space has 52 outcomes, one outcome corresponding to each card.


Relative Occurrence of an outcome. Suppose that a random experiment is performed a certain number of times, say N. Let O be a particular outcome of this experiment, and suppose that O occurs \({N_O}\) times. The relative occurrence of the outcome O will then be \(\frac{{{N_O}}}{N}\). For example, suppose that a coin is tossed 100 times, and H occurs 47 times. Then, the relative occurrence of Heads is \(\frac{{47}}{{100}} = 0.47\).

Equally likely outcomes. Two outcomes of a random experiment are said to be equally likely, if upon performing the experiment a (very) large number of times, the relative occurrences of the two outcomes turn out to be equal. For example, for a perfectly fair coin, the relative occurrences of H and T for a very large number of tosses N will be equal (as N goes to infinity, we will come closer to a perfect equality). Thus, H and T are equally likely outcomes for a perfectly fair coin. Similarly, for an unbiased, unloaded die, each of the six outcomes is equally likely.

Event. For a random experiment, an event is any possible set of outcomes. Examples:

In rolling a die, an event E is defined as “E: 3 is not obtained”. This event has 5 possible outcomes: {1, 2, 4, 5, 6}.

In drawing a card from a deck, an event F is defined as “F: a black card is obtained”. Event F has 26 posible outcomes (13 Spades + 13 Clubs).


Probability. The probability of an event E occurring is actually the same as the value we will obtain for the relative occurrence of E if the experiment is performed a large number (going to infinity) of times. In case all the outcomes are equally likely, we can calculate the probability of E as the ratio of the number of outcomes favorable to E, to the total number of outcomes.
Example 1: Two dice are rolled simultaneously.
(a) What is the sample space for this random experiment?
(b) Define as event E as “E: the sum of the two numbers which show up is 5”. List all the outcomes for this event.
(c) Define an event F as “F: at least one of the two numbers which show up is a 3. List all the outcomes for this event.
(d) Assuming that the dice are fair, calculate the probabilities of E and F.
Solution:
(a) The sample space has 36 possible outcomes, which are listed out explicitly below:
\[\begin{array}{l}\left( {1,1} \right)\,\,\,\left( {2,1} \right)\,\,\,\left( {3,1} \right)\,\,\,\left( {4,1} \right)\,\,\,\left( {5,1} \right)\,\,\,\left( {6,1} \right)\\\left( {1,2} \right)\,\,\,\left( {2,2} \right)\,\,\,\left( {3,2} \right)\,\,\,\left( {4,2} \right)\,\,\,\left( {5,2} \right)\,\,\,\left( {6,2} \right)\\\left( {1,3} \right)\,\,\,\left( {2,3} \right)\,\,\,\left( {3,3} \right)\,\,\,\left( {4,3} \right)\,\,\,\left( {5,3} \right)\,\,\,\left( {6,3} \right)\\\left( {1,4} \right)\,\,\,\left( {2,4} \right)\,\,\,\left( {3,4} \right)\,\,\,\left( {4,4} \right)\,\,\,\left( {5,4} \right)\,\,\,\left( {6,4} \right)\\\left( {1,5} \right)\,\,\,\left( {2,5} \right)\,\,\,\left( {3,5} \right)\,\,\,\left( {4,5} \right)\,\,\,\left( {5,5} \right)\,\,\,\left( {6,5} \right)\\\left( {1,6} \right)\,\,\,\left( {2,6} \right)\,\,\,\left( {3,6} \right)\,\,\,\left( {4,6} \right)\,\,\,\left( {5,6} \right)\,\,\,\left( {6,6} \right)\end{array}\]
(b) The outcomes which are favorable to event E are:
\[\left( {4,1} \right),\,\,\,\left( {3,2} \right),\,\,\,\left( {2,3} \right),\,\,\,\left( {1,4} \right)\]
(c) The outcomes which are favorable to event F are:
\[\begin{array}{l}\left( {1,3} \right)\,\,\,\left( {2,3} \right)\,\,\,\left( {3,3} \right)\,\,\,\left( {4,3} \right)\,\,\,\left( {5,3} \right)\,\,\,\left( {6,3} \right)\\\left( {3,1} \right)\,\,\,\left( {3,2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {3,4} \right)\,\,\,\left( {3,5} \right)\,\,\,\left( {3,6} \right)\end{array}\]
(d) If the outcomes in the sample space are equally likely, we can calculate the probability of any event as the ratio of the number of favorable outcomes to the total number of outcomes. Thus,
\[\begin{align}&P\left( E \right) = \frac{4}{{36}} = \frac{1}{9}\\&P\left( F \right) = \frac{{11}}{{36}}\end{align}\]
Example 2: Seventeen cards numbered from 1 to 17 are put in a box and mixed thoroughly. Manan draws a card from the box. Find the probability of the following events:
(a) E: the card has an odd number
(b) F: the card shows a prime number
(c) G: the number on the card is divisible by 3
(d) H: the number on the card is divisible by 2 and 3 both
Solution: The sample space has 17 possible outcomes. Since the cards have been mixed thoroughly, each outcome is equally likely.
(a) There are 9 odd numbers from 1 to 17. Thus, \(P\left( E \right) = \frac{9}{{17}}\).
(b) There are 7 prime numbers from 1 to 17: {2, 3, 5, 7, 11, 13, 17. Thus, \(P\left( F \right) = \frac{7}{{17}}\).
(c) There are 5 multiples of 3 from 1 to 17: {3, 6, 9, 12, 15}. Thus, \(P\left( G \right) = \frac{5}{{17}}\).
(d) Numbers which are multiples of both 2 and 3 are multiples of 6. There are 2 multiples of 6 from 1 to 17: {6, 12}. Thus, \(P\left( H \right) = \frac{2}{{17}}\).
Example 3: A bag has 43 identical black balls and 7 identical red balls. A ball is drawn at random from this bag.
(a) How many outcomes are there in the sample space?
(b) What is the probability of the ball being black?
Solution:
(a) This problem may present a slight confusion. Since the black balls are identical, and the red balls are identical, you may think that there are only two possible outcomes:

The ball is black

The ball is red
This is alright, but if you define your outcomes in this manner, then the two outcomes are not equally likely, because the number of black balls is higher than the number of red balls. Thus, a black ball is more likely to show up.
When doing probability calculations, we generally like to define our outcomes in a manner so that they are equally likely. In this case, suppose that you mentally number the balls (from 1 to 50). Then, instead of two outcomes, you have 50 outcomes, with each outcome being equally likely (since the balls are identical apart from their numbering you have mentally done).
Thus, we define our sample space so that it has 50 outcomes: 43 corresponding to the 43 black balls, and 7 corresponding to the 7 red balls.
(b) If E is the event that a black ball shows up, then
\[P\left( E \right) = \frac{{\# {\rm{Favorable}}}}{{\# {\rm{Total}}}} = \frac{{43}}{{50}}\]
We were able to use this expression because the 50 outcomes are equally likely.
Example 4: Two dice are rolled simultaneously. It is given that the sum of the two numbers which show up is 8. What is the probability that the product of the two numbers is odd?
Solution: Note that if the fact of the sum of the two numbers being 8 wasn’t specified, then our sample space would have consisted of 36 outcomes. But now, since this fact is already know, our sample space will reduce to the following five outcomes
\[\left( {6,2} \right)\,\,\,\left( {5,3} \right)\,\,\,\left( {4,4} \right)\,\,\,\left( {3,5} \right)\,\,\,\left( {2,6} \right)\]
Define E as the event that the product of the two numbers is odd, given that their sum is 8. Out of these 5 outcomes, the product is odd for 2 outcomes, namely \(\left( {5,3} \right)\) and \(\left( {3,5} \right)\). Thus, 2 of the 5 possible outcomes are favorable to E, so that
\[P\left( E \right) = \frac{2}{5}\]
Example 5: A random experiment consists of tossing a fair coin and rolling an unbiased die together. What is the probability of obtaining Heads on the coin and an even number on the die?
Solution: Let E be the specified event. By the Fundamental Principle of Counting, the sample space consists of \(2 \times 6 = 12\) outcomes, which are listed explicitly below:
\[\begin{array}{l}\left( {H,1} \right),\,\,\left( {H,2} \right),\,\,\left( {H,3} \right),\,\,\left( {H,4} \right),\,\,\left( {H,5} \right),\,\,\left( {H,6} \right)\\\left( {T,1} \right),\,\,\left( {T,2} \right),\,\,\left( {T,3} \right),\,\,\left( {T,4} \right),\,\,\left( {T,5} \right),\,\,\left( {T,6} \right)\end{array}\]
The number of outcomes favorable to E is \(1 \times 3 = 3\), and these three outcomes are:
\[\left( {H,2} \right),\,\,\left( {H,4} \right),\,\,\left( {H,6} \right)\]
Thus, the required probability is
\[P\left( E \right) = \frac{3}{{12}} = \frac{1}{4}\]
We can think of this calculation as follows:
\[P\left( E \right) = \frac{{1 \times 3}}{{2 \times 6}} = \left( {\frac{1}{2}} \right) \times \left( {\frac{3}{6}} \right)\]
The first term in the product, which is \(\left( {\frac{1}{2}} \right)\), is the probability of obtaining Heads on tossing the coin. The second term, which is \(\left( {\frac{3}{6}} \right)\), is the probability of obtaining an even number on rolling the die. Thus, we see that the probability of the two taken together, which is the probability of our event E, is the product of these separate probabilities.
Let us see another similar example.
Example 6: An unbiased die is rolled, and simultaneously, a card is drawn from a wellshuffled deck. The event E is defined as follows:
E: Die does not show 1, Card is a Spades card
Find \(P\left( E \right)\).
Solution: Using the FPC, we see that the sample space consists of \(6 \times 52\) outcomes. The number of favorable outcomes is \(5 \times 13\) (since there are 5 favorable outcomes of the die roll and 13 favorable outcomes of the card draw). Thus,
\[P\left( E \right) = \frac{{5 \times 13}}{{6 \times 52}} = \left( {\frac{5}{6}} \right) \times \left( {\frac{{13}}{{32}}} \right)\]
Once again, this probability is the product of:

The probability of not obtaining a 1 on a die roll, and

The probability of obtained a card of Spades on a card draw
In such scenarios, where two independent random experiments are clubbed together to create a single parent random experiment, the probability of an event of the parent random experiment will be the product of the probabilities of the corresponding events in the two child random experiments.
Let us see yet one more example of this.
Example 7: BagA has 5 red balls and 11 black balls. BagB has 2 red balls and 7 black balls. A random experiment is carried out as follows: a bag is selected at random from these two bags, and then a ball is picked at random from that bag. Find the probability that this ball is red.
Solution: The desired event E (the ball drawn is red) can be split into two subevents:
(a) \({E_1}\): The red ball is drawn from BagA
(b) \({E_2}\): The red ball is drawn from BagB
The two subevents are mutually exclusive, in the sense that for any performance of the random experiment, only one of the two subevents will occur. Let us calculate the probability of each subevent separately.
Event \({E_1}\). This subevent is a combination of the following:

In choosing the bags, BagA is selected
 In drawing the ball from BagA, a red ball is drawn
Thus,
\[\begin{align}&P\left( {{E_1}} \right) = \underbrace {\left( {\frac{1}{2}} \right)}_{\scriptstyle{\rm{Bag  A }}\atop\scriptstyle{\rm{is selected}}} \times \underbrace {\left( {\frac{5}{{5 + 11}}} \right)}_{\scriptstyle{\rm{A\, red\, ball\, is}}\atop\scriptstyle{\rm{drawn}}} = \left( {\frac{1}{2}} \right) \times \left( {\frac{5}{{16}}} \right)\\ \qquad \qquad &\qquad\;\;= \frac{5}{{32}}\end{align}\]
Event \({E_2}\). This subevent is a combination of the following:

In choosing the bags, BagB is selected
 In drawing the ball from BagB, a red ball is drawn
Thus,
\[\begin{align}&P\left( {{E_2}} \right) = \underbrace {\left( {\frac{1}{2}} \right)}_{\scriptstyle{\rm{Bag  B }}\atop\scriptstyle{\rm{is selected}}} \times \underbrace {\left( {\frac{2}{{2 + 7}}} \right)}_{\scriptstyle{\rm{A\, red\, ball \,is}}\atop\scriptstyle{\rm{drawn}}} = \left( {\frac{1}{2}} \right) \times \left( {\frac{2}{9}} \right)\\ \qquad \qquad&\qquad\;\;= \frac{1}{9}\end{align}\]
Now, think carefully: \({E_1}\) and \({E_2}\) are mutually exclusive events, which means if a red ball is drawn, it must have been through an occurrence of \({E_1}\), or through an occurrence of \({E_2}\). Both the events cannot occur simultaneously.
Also, E occurs only if either \({E_1}\) or \({E_2}\) occurs. This means that for a large number of repetitions of the experiment, the relative occurrence of E will be the sum of the relative occurrences of \({E_1}\) and \({E_2}\). Thus,
\[P\left( E \right) = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) = \frac{5}{{32}} + \frac{1}{9} = \frac{{77}}{{288}}\]