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Probability - Event

Grade 11CBSE

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

🔑Concepts

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Sample Space and Event: The Sample Space (SS) is the set of all possible outcomes of a random experiment. An Event (EE) is any subset of the Sample Space (E⊆SE \subseteq S). Visually, if the Sample Space is represented by a rectangular boundary, an Event is a region or a circle within that boundary containing specific outcome points.

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Impossible and Sure Events: An Impossible Event (emptyset\\emptyset) is an event that contains no sample points, such as rolling a 7 on a standard six-sided die. A Sure Event is the Sample Space (SS) itself, containing all possible outcomes. In a Venn diagram, the Sure Event covers the entire rectangular area, while the Impossible Event is an empty set.

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Simple and Compound Events: If an event (EE) has only one sample point of a sample space, it is called a Simple or Elementary Event. If an event has more than one sample point, it is called a Compound Event. For example, in tossing two coins, the event of getting exactly two heads HH\\{HH\\} is simple, while getting at least one head HH,HT,TH\\{HH, HT, TH\\} is compound.

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Complementary Events: For every event AA, there exists another event A′A' (or AcA^c) called the complementary event, which consists of all outcomes in SS that are not in AA. Visually, if AA is a circle inside the Sample Space rectangle, A′A' is the entire region shaded outside that circle but within the rectangle.

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Mutually Exclusive Events: Two events AA and BB are said to be mutually exclusive if the occurrence of any one of them excludes the occurrence of the other (i.e., AcapB=emptysetA \\cap B = \\emptyset). In a Venn diagram, these are represented by two disjoint circles 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 called exhaustive events if their union is equal to the Sample Space SS (E1cupE2cup...cupEn=SE_1 \\cup E_2 \\cup ... \\cup E_n = S). This means that at least one of these events must occur whenever the experiment is performed. Visually, these events together tile or cover the entire surface area of the Sample Space.

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Algebra of Events: New events can be formed using set operations. The Union (AcupBA \\cup B) represents the event 'AA or BB', including outcomes in AA, BB, or both. The Intersection (AcapBA \\cap B) represents the event 'AA and BB', consisting of outcomes common to both. The Difference (A−BA - B) represents outcomes that are in AA but not in BB.

📐Formulae

Probability of an event AA: P(A)=fracn(A)n(S)P(A) = \\frac{n(A)}{n(S)}, where n(A)n(A) is the number of outcomes favorable to AA and n(S)n(S) is the total number of outcomes.

Range of Probability: 0leqP(A)leq10 \\leq P(A) \\leq 1

Probability of Complementary Event: P(A′)=1−P(A)P(A') = 1 - P(A)

Addition Rule for any two events AA and BB: P(AcupB)=P(A)+P(B)−P(AcapB)P(A \\cup B) = P(A) + P(B) - P(A \\cap B)

For Mutually Exclusive Events AA and BB: P(AcupB)=P(A)+P(B)P(A \\cup B) = P(A) + P(B) since P(AcapB)=0P(A \\cap B) = 0

Probability of event 'A but not B': P(A−B)=P(AcapB′)=P(A)−P(AcapB)P(A - B) = P(A \\cap B') = P(A) - P(A \\cap B)

De Morgan's Laws: P((AcupB)′)=P(A′capB′)P((A \\cup B)') = P(A' \\cap B') and P((AcapB)′)=P(A′cupB′)P((A \\cap B)') = P(A' \\cup B')

💡Examples

Problem 1:

Two dice are thrown simultaneously. Find the probability of getting a total sum of 8.

Solution:

  1. The sample space SS for two dice contains 6times6=366 \\times 6 = 36 outcomes. So, n(S)=36n(S) = 36. \n2. Let AA be the event of getting a sum of 8. The outcomes favorable to AA are (2,6),(3,5),(4,4),(5,3),(6,2)\\{(2,6), (3,5), (4,4), (5,3), (6,2)\\}. \n3. The number of favorable outcomes is n(A)=5n(A) = 5. \n4. The probability P(A)=fracn(A)n(S)=frac536P(A) = \\frac{n(A)}{n(S)} = \\frac{5}{36}.

Explanation:

We first determine the total number of possible outcomes for two dice and then list all specific pairs that add up to the required sum.

Problem 2:

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

Solution:

  1. Total cards n(S)=52n(S) = 52. \n2. Let KK be the event of drawing a King. There are 4 Kings, so P(K)=frac452P(K) = \\frac{4}{52}. \n3. Let RR be the event of drawing a Red card. There are 26 Red cards, so P(R)=frac2652P(R) = \\frac{26}{52}. \n4. The intersection KcapRK \\cap R represents Red Kings. There are 2 Red Kings (King of Hearts and King of Diamonds), so P(KcapR)=frac252P(K \\cap R) = \\frac{2}{52}. \n5. Using the addition rule: P(KcupR)=P(K)+P(R)−P(KcapR)P(K \\cup R) = P(K) + P(R) - P(K \\cap R) \n6. P(KcupR)=frac452+frac2652−frac252=frac2852=frac713P(K \\cup R) = \\frac{4}{52} + \\frac{26}{52} - \\frac{2}{52} = \\frac{28}{52} = \\frac{7}{13}.

Explanation:

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