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Probability - Classical Definition of Probability

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Random Experiment and Outcomes: A random experiment is a process where the result cannot be predicted with absolute certainty. The possible results of such an experiment are called outcomes. Visually, you can imagine a tree diagram where the starting point is the action (like flipping a coin) and the branches represent the possible outcomes (Heads or Tails).

Sample Space: The set of all possible outcomes of a random experiment is called the sample space, denoted by SS. Imagine a rectangular 'Venn Diagram' frame containing several distinct points; each point represents a unique outcome, and the entire area inside the frame is the Sample Space.

Equally Likely Outcomes: Outcomes are said to be equally likely if none of them is expected to occur in preference to the others. For example, in a fair six-sided die, each face from 11 to 66 has the exact same physical area and weight, ensuring each number has a 16\frac{1}{6} chance of appearing.

Theoretical Probability: Also known as Classical Probability, it is calculated based on the assumption that all outcomes are equally likely. It is the ratio of the number of favorable outcomes to the total number of outcomes. Visually, if the sample space is a pie chart, the probability of an event is the 'slice' of the pie that represents the favorable outcomes.

Range of Probability: The probability of an event EE is always a value such that 0P(E)10 \le P(E) \le 1. On a probability number line, 00 indicates an 'Impossible Event' (like rolling a 77 on a standard die) and 11 indicates a 'Sure Event' (like rolling a number less than 77 on a standard die).

Complementary Events: For every event EE, there exists an event 'not EE' (denoted as Eˉ\bar{E} or EE') which represents EE not occurring. If you visualize a circle representing the Sample Space, and a smaller circle inside it is EE, then everything outside the small circle but inside the large circle is Eˉ\bar{E}.

Elementary Events: An event that consists of only one outcome of the experiment is called an elementary event. A key property is that the sum of the probabilities of all the elementary events of an experiment is exactly 11.

Standard Deck of Cards: A standard deck contains 5252 cards, split into 44 suits: Spades (♠), Hearts (♥), Diamonds (♦), and Clubs (♣). There are 2626 red cards (Hearts and Diamonds) and 2626 black cards (Spades and Clubs). Each suit has 1313 cards: Ace, 2,3,4,5,6,7,8,9,102, 3, 4, 5, 6, 7, 8, 9, 10, and three 'face cards' (King, Queen, Jack).

📐Formulae

P(E)=Number of outcomes favorable to ENumber of all possible outcomes of the experimentP(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Number of all possible outcomes of the experiment}}

0P(E)10 \le P(E) \le 1

P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1 or P(not E)=1P(E)P(\text{not } E) = 1 - P(E)

P(ϕ)=0P(\phi) = 0 (Probability of an impossible event)

P(S)=1P(S) = 1 (Probability of a sure event)

Sum of probabilities of all elementary events=P(Ei)=1\text{Sum of probabilities of all elementary events} = \sum P(E_i) = 1

💡Examples

Problem 1:

A die is thrown once. What is the probability of getting (i) a prime number and (ii) a number lying between 22 and 66?

Solution:

When a die is thrown, the total number of possible outcomes is 66, which are {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}. \n (i) Prime numbers on a die are 2,3,2, 3, and 55. So, the number of outcomes favorable to prime numbers is 33. \n P(Prime Number)=36=12P(\text{Prime Number}) = \frac{3}{6} = \frac{1}{2}. \n (ii) Numbers lying between 22 and 66 are 3,4,3, 4, and 55. So, the number of outcomes favorable is 33. \n P(Number between 2 and 6)=36=12P(\text{Number between 2 and 6}) = \frac{3}{6} = \frac{1}{2}.

Explanation:

First, identify the total outcomes in the sample space. Then, list the specific outcomes that satisfy the given condition to find the numerator for the probability formula.

Problem 2:

Two coins are tossed simultaneously. Find the probability of getting at least one head.

Solution:

When two coins are tossed, the sample space SS is {HH,HT,TH,TT}\{HH, HT, TH, TT\}. \n The total number of possible outcomes is 44. \n Let EE be the event of getting 'at least one head'. The outcomes favorable to EE are {HH,HT,TH}\{HH, HT, TH\}. \n The number of favorable outcomes is 33. \n Therefore, P(E)=Number of favorable outcomesTotal number of outcomes=34P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}.

Explanation:

In multi-stage experiments, listing the sample space carefully is crucial. 'At least one' means one or more, so we include all outcomes except the one with no heads (TT).