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Probability - Probability of an event

Grade 11CBSE

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

🔑Concepts

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Random Experiment and Sample Space: A random experiment is an action where the outcome cannot be predicted with certainty, such as tossing a coin. The Sample Space (SS) is the set of all possible outcomes. Visually, SS is often represented as a large rectangular box (a Venn diagram boundary) containing points that represent every possible outcome.

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Events as Subsets: An event (EE) is a subset of the Sample Space (SS). In a visual Venn diagram, an event is depicted as a closed loop (like a circle or oval) inside the SS rectangle. If an actual outcome falls within this loop, we say the event has occurred.

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Mutually Exclusive Events: Two events AA and BB are mutually exclusive if they cannot happen at the same time, meaning A∩B=ϕA \cap B = \phi. Visually, these are represented as two disjoint circles within the sample space that do not overlap or share any common area.

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Exhaustive Events: Events E1,E2,...,EnE_1, E_2, ..., E_n are exhaustive if their union covers the entire sample space, i.e., E1∪E2∪...∪En=SE_1 \cup E_2 \cup ... \cup E_n = S. Visually, if you combined the areas of all these event shapes, they would perfectly fill the entire rectangle of the sample space.

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Impossible and Sure Events: An impossible event is represented by the empty set ϕ\phi and has a probability of 00. A sure event is the entire sample space SS and has a probability of 11. On a probability scale (a number line from 00 to 11), the impossible event is at the far left and the sure event is at the far right.

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Complement of an Event: The complement of an event AA, denoted as A′A' or AcA^c, consists of all outcomes in SS that are not in AA. Visually, if AA is a circle inside the rectangle SS, the complement A′A' is the entire shaded region inside the rectangle but outside the circle AA.

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Equally Likely Outcomes: Outcomes are equally likely if none is preferred over the other. If a sample space has nn outcomes, each has a probability of 1n\frac{1}{n}. This can be visualized as a pie chart where every slice is of equal size and angle.

📐Formulae

P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}

0≤P(E)≤10 \le P(E) \le 1

P(S)=1 and P(ϕ)=0P(S) = 1 \text{ and } P(\phi) = 0

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A′)=1−P(A)P(A') = 1 - P(A)

P(A∪B)=P(A)+P(B) (if A,B are mutually exclusive)P(A \cup B) = P(A) + P(B) \text{ (if } A, B \text{ are mutually exclusive)}

P(A−B)=P(A∩B′)=P(A)−P(A∩B)P(A - B) = P(A \cap B') = P(A) - P(A \cap B)

💡Examples

Problem 1:

Two fair dice are rolled simultaneously. Find the probability that the sum of the numbers appearing on the top faces is 9.

Solution:

  1. The total number of outcomes when two dice are rolled is n(S)=6×6=36n(S) = 6 \times 6 = 36. \n2. Let EE be the event that the sum is 9. The outcomes favorable to EE are {(3,6),(4,5),(5,4),(6,3)}\{(3, 6), (4, 5), (5, 4), (6, 3)\}. \n3. Therefore, the number of favorable outcomes is n(E)=4n(E) = 4. \n4. The probability is P(E)=n(E)n(S)=436=19P(E) = \frac{n(E)}{n(S)} = \frac{4}{36} = \frac{1}{9}.

Explanation:

This problem uses the classical definition of probability. We first identify the size of the sample space and then list the specific pairs that satisfy the condition (sum = 9) to find the probability.

Problem 2:

A card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card is either a Red card or a King?

Solution:

  1. Total cards n(S)=52n(S) = 52. \n2. Let AA be the event of drawing a Red card. There are 26 red cards, so n(A)=26n(A) = 26. Thus, P(A)=2652P(A) = \frac{26}{52}. \n3. Let BB be the event of drawing a King. There are 4 kings in a deck, so n(B)=4n(B) = 4. Thus, P(B)=452P(B) = \frac{4}{52}. \n4. A∩BA \cap B is the event of drawing a Red King. There are 2 red kings (Hearts and Diamonds), so n(A∩B)=2n(A \cap B) = 2. Thus, P(A∩B)=252P(A \cap B) = \frac{2}{52}. \n5. Using the addition rule: P(A∪B)=P(A)+P(B)−P(A∩B)=2652+452−252=2852=713P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}.

Explanation:

Since a card can be both Red and a King, these events are not mutually exclusive. We must use the general addition formula and subtract the intersection to avoid double-counting the red kings.