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Chapter 15 Probability (Concepts)
Welcome to this crucial chapter where we solidify our understanding of Probability, transitioning significantly from the experimental (or empirical) approach often emphasized in earlier studies towards a more theoretical and foundational perspective. While experimental probability relies on observed frequencies from trials, this chapter focuses predominantly on the Classical (or Theoretical) Definition of Probability. This approach allows us to calculate probabilities based on logical reasoning and the inherent structure of the experiment, assuming a fundamental condition of fairness or symmetry.
The classical definition hinges on the assumption that all possible outcomes of a random experiment are equally likely. This means that each individual outcome has the same chance of occurring (e.g., a fair coin is equally likely to land heads or tails; each face of a fair standard die has an equal chance of appearing). Under this assumption, the theoretical probability of a specific event E occurring, denoted as $P(E)$, is defined as the ratio of the number of outcomes that are favourable to the event E, to the total number of possible outcomes in the experiment's sample space. Mathematically, this is expressed as: $$ \mathbf{P(E) = \frac{\text{Number of outcomes favourable to E}}{\text{Total number of possible outcomes}}} $$ We also consider an elementary event, which is an event consisting of only a single outcome in the sample space. A fundamental property is that the sum of the probabilities of all the elementary events associated with an experiment must always equal 1.
Several key properties and concepts governing probability are rigorously established:
- The probability of any event E must always fall within the range of 0 to 1, inclusive. That is, $\mathbf{0 \le P(E) \le 1}$. A probability can never be negative or greater than 1.
- An event that is impossible to occur (it has no favourable outcomes) is called an impossible event, and its probability is exactly 0.
- An event that is guaranteed to occur (all possible outcomes are favourable outcomes) is called a sure event or certain event, and its probability is exactly 1.
Another vital concept is that of Complementary Events. For any event E, the event 'not E' represents the outcome that E does not occur. This complementary event is often denoted as $E'$ or $\overline{E}$. Since an event either happens or does not happen, the sum of their probabilities must be 1. This gives us the crucial relationship: $P(E) + P(\text{not } E) = 1$. This formula is extremely useful, as it allows us to calculate the probability of an event not happening if we know the probability that it does happen, and vice versa: $P(\text{not } E) = 1 - P(E)$.
The practical application of this theoretical framework involves calculating probabilities for various standard random experiments. Common scenarios include:
- Tossing one fair coin ($H, T$) or multiple fair coins (e.g., two coins: $HH, HT, TH, TT$).
- Rolling one fair standard six-sided die (outcomes $1, 2, 3, 4, 5, 6$) or multiple dice (understanding the 36 equally likely outcomes when rolling two dice).
- Drawing a single card from a standard, well-shuffled deck of 52 playing cards. This requires familiarity with the deck's structure: 4 suits (Hearts $\color{red}{\heartsuit}$, Diamonds $\color{red}{\diamondsuit}$, Clubs $\clubsuit$, Spades $\spadesuit$), 2 colors (Red/Black), 13 ranks per suit (A, 2-10, J, Q, K), including 12 face cards (J, Q, K) in total and 4 Aces.
- Selecting an object at random from a collection containing items of different types (e.g., choosing a colored ball from an urn or a marble from a bag).
Success in solving probability problems hinges on accurately identifying two key quantities: the total number of possible, equally likely outcomes (the size of the sample space) and the specific number of outcomes that satisfy the conditions of the event in question (favourable outcomes). The focus is on systematic counting and applying the fundamental definition $P(E) = \frac{\text{Favourable}}{\text{Total}}$.
Probability - Experimental and Theoretical Approaches
In our daily experiences, we frequently encounter situations where the outcome is uncertain. We use terms like 'likely', 'unlikely', 'possibly', 'chance', 'probably', etc., to express this uncertainty. Probability is the branch of mathematics that provides a way to measure and quantify this uncertainty, giving us a numerical value for the likelihood that a specific event will occur.
Random Experiment, Trial, Outcome, and Event
To discuss probability mathematically, we need to be precise about the procedures and their results:
- A random experiment is an experiment or a process with the following characteristics:
- All possible outcomes of the experiment are known in advance.
- The result of a specific trial cannot be predicted with certainty beforehand.
- A trial is a single performance or execution of a random experiment. Example: Tossing a coin once is a trial. Rolling a die one time is a trial. Drawing one card is a trial.
- An outcome is a possible result of a single trial of a random experiment. Examples: Getting a 'Head' when tossing a coin is an outcome. Getting a '3' when rolling a die is an outcome.
- An event is a specific outcome or a collection of one or more outcomes from a random experiment that we are interested in. An event is a subset of the sample space (the set of all possible outcomes). Examples: Getting a 'Head' when tossing a coin is an event (it consists of a single outcome). Getting an even number when rolling a die is an event (it consists of the outcomes {2, 4, 6}). Getting 'at least one Head' when tossing two coins is an event ({HH, HT, TH}).
Experimental Probability (Empirical Probability)
Experimental probability, also known as empirical probability or observed probability, is determined by actually performing a random experiment a number of times and recording the outcomes. It is calculated based on the observed frequency of occurrence of an event during these trials.
The formula for calculating the experimental probability of an event E is:
$P(E) = \frac{\text{Number of times the event E occurred}}{\text{Total number of trials}}$
... (1)
This type of probability is based on observed results and is an estimation of the true probability. The reliability of this estimation generally increases as the total number of trials increases. In Class 9, the focus was primarily on this approach.
Example 1. A coin is tossed 100 times with the following frequencies:
Heads: 45 times
Tails: 55 times
Find the experimental probability of getting a Head and the experimental probability of getting a Tail.
Answer:
To Find:
The experimental probability of getting a Head.
The experimental probability of getting a Tail.
Given:
Total number of trials (coin tosses) = 100.
Number of Heads (Frequency of Head) = 45 times.
Number of Tails (Frequency of Tail) = 55 times.
Solution:
We use the formula for experimental probability (Formula 1).
Let $E_{\text{Head}}$ be the event of getting a Head.
$P(E_{\text{Head}}) = \frac{\text{Number of Heads}}{\text{Total number of tosses}}$
Substitute the given values:
$P(E_{\text{Head}}) = \frac{45}{100} = 0.45$
... (1)
Let $E_{\text{Tail}}$ be the event of getting a Tail.
$P(E_{\text{Tail}}) = \frac{\text{Number of Tails}}{\text{Total number of tosses}}$
Substitute the given values:
$P(E_{\text{Tail}}) = \frac{55}{100} = 0.55$
... (2)
Answer: The experimental probability of getting a Head is 0.45, and the experimental probability of getting a Tail is 0.55.
Note that the sum of the probabilities of getting a Head or a Tail is $0.45 + 0.55 = 1$. These two events (Head and Tail) are the only possible outcomes and are mutually exclusive in a single toss.
Theoretical Probability (Classical Probability)
Theoretical probability, also referred to as classical probability, is based on the analysis of all possible outcomes of a random experiment, assuming that all these outcomes are equally likely to occur. This means that each outcome has the same chance of happening. Theoretical probability is calculated based on reasoning and knowledge of the experiment's setup, without needing to perform actual trials.
For experiments where all outcomes are equally likely, the theoretical probability of an event E is defined as:
$P(E) = \frac{\text{Number of outcomes favourable to E}}{\text{Total number of all possible outcomes}}$
... (4)
Here, 'Total number of all possible outcomes' is the total count of outcomes in the sample space, and 'Number of outcomes favourable to E' is the count of outcomes in the sample space that satisfy the definition of the event E.
Example: For a fair coin toss, the possible outcomes are Head and Tail. There are 2 total possible outcomes, and they are equally likely. For the event 'Getting a Head', there is 1 favourable outcome (Head). Theoretical $P(\text{Head}) = \frac{1}{2}$. For the event 'Getting a Tail', there is 1 favourable outcome (Tail). Theoretical $P(\text{Tail}) = \frac{1}{2}$.
The relationship between experimental and theoretical probability is that as the number of trials in an experiment increases indefinitely, the experimental probability of an event tends to approach its theoretical probability. This concept is formalised in the Law of Large Numbers.
Theoretical Probability
Theoretical probability, also referred to as classical probability, is a mathematical approach to probability that is based on reasoning about the possible outcomes of an experiment, rather than conducting actual trials. It is applicable in situations where all possible outcomes of the experiment are considered equally likely.
Equally Likely Outcomes
The outcomes of a random experiment are considered equally likely if each outcome has the same chance of occurring as any other outcome. In such cases, there is no inherent bias towards any particular outcome.
Examples of experiments with equally likely outcomes:
- Tossing a fair coin (Head and Tail are equally likely).
- Rolling a fair standard six-sided die (each face 1, 2, 3, 4, 5, 6 is equally likely to appear).
- Drawing any single card from a well-shuffled standard deck of cards (each card is equally likely to be drawn).
Examples of experiments with outcomes that are NOT equally likely:
- Tossing a biased coin.
- Rolling a loaded die.
- Drawing a ball from a bag containing an unequal number of balls of different colors (e.g., 10 red balls and 2 blue balls).
Formula for Theoretical Probability
For a random experiment where all possible outcomes are equally likely, the theoretical probability of an event E, denoted by P(E), is calculated using the following formula:
Theoretical Probability Formula. For an experiment with equally likely outcomes, the probability of an event E is:
$P(E) = \frac{\text{Number of outcomes favourable to E}}{\text{Total number of all possible outcomes}}$
Here:
- 'Total number of all possible outcomes' is the size of the sample space (the total count of all possible results of the experiment).
- 'Number of outcomes favourable to E' is the count of outcomes within the sample space that satisfy the condition(s) defining the event E.
This formula is applicable when the total number of possible outcomes is finite and can be enumerated, and each outcome is equally probable.
Elementary Event
An elementary event (or simple event) is an event that consists of only one single outcome of the experiment. Each individual outcome in the sample space is an elementary event.
Example: In rolling a standard die, getting a 1 is an elementary event. Getting a 2 is another elementary event, and so on. There are 6 elementary events in this experiment.
The sum of the probabilities of all the elementary events of an experiment is always equal to 1. This makes sense because one of the elementary outcomes is certain to occur in any given trial.
Impossible Event and Sure Event
Based on the number of favourable outcomes relative to the total number of possible outcomes, we define two special types of events:
- An impossible event is an event that cannot occur in any trial of the experiment. The number of outcomes favourable to an impossible event is 0. Example: Getting a 7 when rolling a standard six-sided die is an impossible event. The probability of an impossible event is $P(\text{Impossible Event}) = \frac{0}{\text{Total Outcomes}} = 0$.
- A sure event (or certain event) is an event that is certain to occur in every trial of the experiment. The number of outcomes favourable to a sure event is equal to the total number of all possible outcomes. Example: Getting a number less than 7 when rolling a standard six-sided die is a sure event (as all possible outcomes 1, 2, 3, 4, 5, 6 are less than 7). The probability of a sure event is $P(\text{Sure Event}) = \frac{\text{Total Outcomes}}{\text{Total Outcomes}} = 1$.
Range of Probability
From the definitions of probability, the probability of any event E is always a number between 0 and 1, inclusive. It cannot be negative, and it cannot be greater than 1.
$0 \leq P(E) \leq 1$
Complementary Event
For any event E, the event that E does not occur is called the complementary event of E. It is denoted by E' or $\overline{E}$. An event E and its complementary event E' are mutually exclusive (they cannot happen at the same time) and exhaustive (together they cover all possible outcomes of the experiment).
The sum of the probabilities of an event and its complementary event is always 1. This is because either the event E happens or it does not happen (E' happens), covering all possibilities with no overlap.
Complementary Event Probability Rule. For any event E,
$P(E) + P(E') = 1$
This relationship can be rearranged to find the probability of the complementary event:
$P(E') = 1 - P(E)$
This rule is very useful when it is easier to calculate the probability of the complementary event than the event itself.
Probability Related to Tossing of Coin(s)
Tossing a coin (or multiple coins) is a fundamental example of a random experiment used in probability theory. For a fair coin, the outcomes of a toss are considered equally likely, making it suitable for illustrating theoretical probability concepts.
Single Coin Toss
When a single fair coin is tossed once, there are only two possible outcomes: Head (H) and Tail (T).
The sample space is S = {H, T}.
The total number of possible outcomes is 2. Since the coin is fair, these outcomes are equally likely.
- The event of getting a Head is {H}. The number of favourable outcomes is 1. The theoretical probability is:
$P(\text{Head}) = \frac{\text{Number of favourable outcomes}}{\text{Total outcomes}} = \frac{1}{2}$
... (1)
- The event of getting a Tail is {T}. The number of favourable outcomes is 1. The theoretical probability is:
$P(\text{Tail}) = \frac{\text{Number of favourable outcomes}}{\text{Total outcomes}} = \frac{1}{2}$
... (2)
Note that P(Head) + P(Tail) = $\frac{1}{2} + \frac{1}{2} = 1$, as Head and Tail are the only possible elementary events.
Tossing Two Coins Simultaneously
When two fair coins are tossed simultaneously, or when a single fair coin is tossed twice, each outcome is an ordered pair.
The possible outcomes are:
- HH (Head on the first, Head on the second)
- HT (Head on the first, Tail on the second)
- TH (Tail on the first, Head on the second)
- TT (Tail on the first, Tail on the second)
The sample space is S = {HH, HT, TH, TT}. The total number of equally likely outcomes is 4.
Example 1. Two coins are tossed simultaneously. Find the probability of:
(a) Getting exactly one Head.
(b) Getting at least one Head.
(c) Getting no Head.
Answer:
To Find:
The probabilities of the given events.
Given:
Two fair coins are tossed simultaneously.
Solution:
The sample space is S = {HH, HT, TH, TT}. Total number of outcomes = 4.
(a) Event of getting exactly one Head.
Let $E_1$ be this event. The outcomes with exactly one Head are {HT, TH}.
Number of favourable outcomes = 2.
$P(E_1) = \frac{2}{4} = \frac{1}{2}$
... (1)
(b) Event of getting at least one Head.
Let $E_2$ be this event. 'At least one Head' means one Head or two Heads. The favourable outcomes are {HH, HT, TH}.
Number of favourable outcomes = 3.
$P(E_2) = \frac{3}{4}$
... (2)
(c) Event of getting no Head.
Let $E_3$ be this event. 'No Head' means getting Tails on both coins. The favourable outcome is {TT}.
Number of favourable outcomes = 1.
$P(E_3) = \frac{1}{4}$
... (3)
Note: The event 'getting no Head' ($E_3$) is the complementary event to 'getting at least one Head' ($E_2$). We can check this: $P(E_2) + P(E_3) \ $$ = \frac{3}{4} + \frac{1}{4} = 1$.
Answer: The probabilities are: (a) $\frac{1}{2}$, (b) $\frac{3}{4}$, and (c) $\frac{1}{4}$.
Tossing Three Coins Simultaneously
When three fair coins are tossed simultaneously, the sample space expands. There are $2 \times 2 \times 2 = 2^3 = 8$ possible outcomes.
The sample space is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
The total number of equally likely outcomes is 8.
Example 2. Three coins are tossed simultaneously. Find the probability of getting at most one Head.
Answer:
To Find:
The probability of getting at most one Head.
Given:
Three fair coins are tossed simultaneously.
Solution:
The sample space has 8 possible outcomes: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
The event 'getting at most one Head' means getting zero Heads OR getting exactly one Head.
- Outcomes with zero Heads: {TTT}
- Outcomes with exactly one Head: {HTT, THT, TTH}
The set of favourable outcomes is E = {TTT, HTT, THT, TTH}.
The number of favourable outcomes is 4.
$P(\text{At most one Head}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
$P(\text{At most one Head}) = \frac{4}{8} = \frac{1}{2}$
... (1)
Answer: The probability of getting at most one Head is $\frac{1}{2}$.
Probability Related to a Pack of 52 Cards
Problems involving drawing cards from a standard deck are common in theoretical probability, as drawing any specific card from a well-shuffled deck is considered an equally likely event. To calculate probabilities in such scenarios, it is essential to understand the composition and structure of a standard deck of playing cards.
Composition of a Standard Deck of 52 Playing Cards
A standard deck of playing cards consists of 52 cards. This deck is divided into suits, colors, and ranks.
- Total Cards: 52.
- Colors: The deck is divided into two colors.
- Red Cards (26): All Hearts ($\heartsuit$) and Diamonds ($\diamondsuit$).
- Black Cards (26): All Clubs ($\clubsuit$) and Spades ($\spadesuit$).
- Suits: There are 4 suits, with 13 cards in each suit.
- Hearts ($\heartsuit$) - Red
- Diamonds ($\diamondsuit$) - Red
- Clubs ($\clubsuit$) - Black
- Spades ($\spadesuit$) - Black
- Ranks: Each suit has 13 cards of different ranks.
- Ace (A): There are 4 Aces in a deck (one for each suit).
- Number Cards (2-10): There are 9 number cards in each suit. Total number cards = $9 \times 4 = 36$.
- Face Cards (J, Q, K): These are the Jack, Queen, and King.
- There are 3 face cards in each suit.
- Total face cards = $3 \times 4 = 12$. (4 Jacks, 4 Queens, 4 Kings).
When a card is drawn from a well-shuffled deck, the total number of equally likely outcomes is 52.
Example 1. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(a) A King
(b) A face card
(c) A red face card
(d) The Queen of Diamonds
(e) A Spade
Answer:
To Find:
The probabilities of the given events.
Given:
A single card is drawn from a well-shuffled deck of 52 cards.
Solution:
The total number of possible outcomes is 52. We use the formula $P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.
(a) Event of getting a King.
There are 4 Kings in a deck.
Number of favourable outcomes = 4.
$P(\text{King}) = \frac{4}{52} = \frac{1}{13}$
... (1)
(b) Event of getting a face card.
There are 12 face cards in a deck (4 Jacks, 4 Queens, 4 Kings).
Number of favourable outcomes = 12.
$P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13}$
... (2)
(c) Event of getting a red face card.
There are 6 red face cards (J, Q, K of Hearts and J, Q, K of Diamonds).
Number of favourable outcomes = 6.
$P(\text{Red Face Card}) = \frac{6}{52} = \frac{3}{26}$
... (3)
(d) Event of getting the Queen of Diamonds.
There is only one Queen of Diamonds in a deck.
Number of favourable outcomes = 1.
$P(\text{Queen of Diamonds}) = \frac{1}{52}$
... (4)
(e) Event of getting a Spade.
There are 13 cards in the suit of Spades.
Number of favourable outcomes = 13.
$P(\text{Spade}) = \frac{13}{52} = \frac{1}{4}$
... (5)
Answer: The probabilities are: (a) $\frac{1}{13}$, (b) $\frac{3}{13}$, (c) $\frac{3}{26}$, (d) $\frac{1}{52}$, (e) $\frac{1}{4}$.
Probability Related to Throwing of Dice
Throwing dice is another classic example of a random experiment in probability. Assuming the dice are fair, each outcome is equally likely, making these problems ideal for illustrating theoretical probability.
Single Die Roll
When a single fair six-sided die is rolled once, the possible outcomes are the numbers on its faces: {1, 2, 3, 4, 5, 6}.
The sample space is S = {1, 2, 3, 4, 5, 6}.
The total number of equally likely outcomes is 6.
Example 1. A fair die is rolled once. Find the probability of getting:
(a) An even number
(b) A prime number
(c) A number greater than 4
Answer:
To Find:
The probabilities of the given events.
Given:
A single roll of a fair six-sided die.
Solution:
The sample space S = {1, 2, 3, 4, 5, 6}, and the total number of outcomes is 6.
(a) Event of getting an even number.
The favourable outcomes are {2, 4, 6}. Number of favourable outcomes = 3.
$P(\text{Even Number}) = \frac{3}{6} = \frac{1}{2}$
... (1)
(b) Event of getting a prime number.
The prime numbers in the sample space are {2, 3, 5}. (Note: 1 is not a prime number).
Number of favourable outcomes = 3.
$P(\text{Prime Number}) = \frac{3}{6} = \frac{1}{2}$
... (2)
(c) Event of getting a number greater than 4.
The favourable outcomes are {5, 6}.
Number of favourable outcomes = 2.
$P(\text{Number > 4}) = \frac{2}{6} = \frac{1}{3}$
... (3)
Answer: The probabilities are: (a) $\frac{1}{2}$, (b) $\frac{1}{2}$, and (c) $\frac{1}{3}$.
Throwing Two Dice Simultaneously
When two fair six-sided dice are rolled together, the outcome is an ordered pair $(d_1, d_2)$, where $d_1$ is the result of the first die and $d_2$ is the result of the second. The total number of possible outcomes is $6 \times 6 = 36$.
The sample space can be visualized in a table:
| (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) |
Example 2. Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is 7.
Answer:
To Find:
The probability that the sum of the numbers on the two dice is 7.
Solution:
The total number of possible outcomes is 36.
We need to find the outcomes where the sum of the two dice is 7. Let's list them:
The favourable outcomes are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1).
The number of favourable outcomes is 6.
Using the probability formula:
$P(\text{Sum is 7}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36}$
$P(\text{Sum is 7}) = \frac{1}{6}$
... (1)
Answer: The probability that the sum of the numbers is 7 is $\frac{1}{6}$.