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Statistics and Probability - Probability (Axiomatic Approach)

Grade 11ICSE

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 a process where the result is uncertain. The sample space SS is the set of all possible outcomes. Visually, imagine a large rectangle (Venn diagram) representing the entire universe of possibilities, where every distinct point inside represents a unique outcome of the experiment.

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Event as a Subset: An event AA is a subset of the sample space SS. In a visual representation, an event is depicted as a closed loop or circle within the rectangular sample space. If an outcome falls inside this loop, the event AA has occurred.

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Axiomatic Definition of Probability: For a sample space S={w1,w2,…,wn}S = \{w_1, w_2, \dots, w_n\}, the probability PP is a function that assigns a real number to each event such that: (i) 0≤P(A)≤10 \le P(A) \le 1 for any event AA, (ii) P(S)=1P(S) = 1 representing the sure event, and (iii) for mutually exclusive events, the probability of their union is the sum of their individual probabilities.

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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=∅A \cap B = \emptyset. Visually, this is shown as two separate circles within the sample space rectangle that do not overlap or touch at any point.

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Exhaustive Events: Events E1,E2,…,EnE_1, E_2, \dots, E_n are exhaustive if their union equals the sample space SS. Visually, if you look at the sample space rectangle, it would be entirely covered by the regions representing these events, leaving no empty space within the rectangle.

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Complementary Events: 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, A′A' is the entire shaded area of the rectangle that lies outside that circle.

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Sure and Impossible Events: The sample space SS itself is the sure event, as one of its outcomes must occur. The empty set ∅\emptyset is the impossible event, containing no outcomes. Visually, the sure event is the entire rectangle, while the impossible event has no region or area assigned to it.

📐Formulae

Classical Probability: P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}

Probability Range: 0≤P(E)≤10 \le P(E) \le 1

Complementary Rule: P(A′)=1−P(A)P(A') = 1 - P(A)

Addition Theorem for Two Events: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Mutually Exclusive Events: P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B) because P(A∩B)=0P(A \cap B) = 0

Difference of Events: P(A−B)=P(A∩B′)=P(A)−P(A∩B)P(A - B) = P(A \cap B') = P(A) - P(A \cap B)

De Morgan's Laws in Probability: P(A′∩B′)=P((A∪B)′)=1−P(A∪B)P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) and P(A′∪B′)=P((A∩B)′)=1−P(A∩B)P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)

Addition Theorem for Three Events: P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)

💡Examples

Problem 1:

In a random experiment, let AA and BB be two events such that P(A)=0.54P(A) = 0.54, P(B)=0.69P(B) = 0.69, and P(A∩B)=0.35P(A \cap B) = 0.35. Find (i) P(A∪B)P(A \cup B), (ii) P(A′∩B′)P(A' \cap B'), and (iii) P(A∩B′)P(A \cap B').

Solution:

Step 1: To find P(A∪B)P(A \cup B), use the Addition Theorem: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) P(A∪B)=0.54+0.69−0.35=0.88P(A \cup B) = 0.54 + 0.69 - 0.35 = 0.88

Step 2: To find P(A′∩B′)P(A' \cap B'), apply De Morgan's Law: P(A′∩B′)=P((A∪B)′)P(A' \cap B') = P((A \cup B)') P(A′∩B′)=1−P(A∪B)=1−0.88=0.12P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.88 = 0.12

Step 3: To find P(A∩B′)P(A \cap B'), use the difference formula: P(A∩B′)=P(A)−P(A∩B)P(A \cap B') = P(A) - P(A \cap B) P(A∩B′)=0.54−0.35=0.19P(A \cap B') = 0.54 - 0.35 = 0.19

Explanation:

The problem applies the basic axiomatic laws of probability. We use the addition theorem for the union, De Morgan's law for the intersection of complements, and the subtraction property for the occurrence of only one event.

Problem 2:

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

Solution:

Step 1: Define the sample space n(S)=52n(S) = 52. Step 2: Let RR be the event of drawing a red card. There are 26 red cards, so n(R)=26n(R) = 26 and P(R)=2652=12P(R) = \frac{26}{52} = \frac{1}{2}. Step 3: Let KK be the event of drawing a King. There are 4 Kings in a deck, so n(K)=4n(K) = 4 and P(K)=452=113P(K) = \frac{4}{52} = \frac{1}{13}. Step 4: Identify the intersection R∩KR \cap K (Red Kings). There are 2 red Kings (King of Hearts and King of Diamonds), so n(R∩K)=2n(R \cap K) = 2 and P(R∩K)=252=126P(R \cap K) = \frac{2}{52} = \frac{1}{26}. Step 5: Apply the Addition Theorem: P(R∪K)=P(R)+P(K)−P(R∩K)P(R \cup K) = P(R) + P(K) - P(R \cap K) P(R∪K)=2652+452−252=2852=713P(R \cup K) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}

Explanation:

Since the events 'Red card' and 'King' are not mutually exclusive (as red kings exist), we must subtract the probability of the intersection to avoid double-counting.