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Probability - Occurrence of an event

Grade 11CBSE

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

🔑Concepts

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The Sample Space SS is the set of all possible outcomes of a random experiment.

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An Event EE is defined as any subset of the sample space SS. If the outcome ω\omega of an experiment is such that ω∈E\omega \in E, we say the event EE has occurred.

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An Impossible Event is represented by the empty set ϕ\phi, and a Sure Event is the entire sample space SS.

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A Simple Event contains only one sample point. If an event has more than one sample point, it is called a Compound Event.

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The Complementary Event of AA, denoted by A′A' or Aˉ\bar{A}, represents the event 'not AA' and consists of all outcomes in SS that are not in AA.

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Two events AA and BB are Mutually Exclusive if they cannot occur simultaneously, meaning A∩B=ϕA \cap B = \phi.

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Events E1,E2,…,EnE_1, E_2, \dots, E_n are Exhaustive Events if their union constitutes the entire sample space, i.e., E1∪E2∪⋯∪En=SE_1 \cup E_2 \cup \dots \cup E_n = S.

📐Formulae

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

0≤P(E)≤10 \le P(E) \le 1

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

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

💡Examples

Problem 1:

A fair die is rolled. Let EE be the event 'the number appearing is a multiple of 3' and FF be the event 'the number appearing is even'. Find P(E∪F)P(E \cup F).

Solution:

The sample space is S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}, so n(S)=6n(S) = 6. Event E={3,6}  ⟹  n(E)=2E = \{3, 6\} \implies n(E) = 2. Event F={2,4,6}  ⟹  n(F)=3F = \{2, 4, 6\} \implies n(F) = 3. The intersection E∩F={6}  ⟹  n(E∩F)=1E \cap F = \{6\} \implies n(E \cap F) = 1. Using the addition rule: P(E∪F)=P(E)+P(F)−P(E∩F)P(E \cup F) = P(E) + P(F) - P(E \cap F) P(E∪F)=26+36−16=46=23P(E \cup F) = \frac{2}{6} + \frac{3}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}

Explanation:

We first identify the outcomes for each event within the sample space of a die. Since the events are not mutually exclusive (the outcome 6 is common to both), we use the general addition formula to find the probability of either event occurring.

Problem 2:

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

Solution:

The sample space S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}, so n(S)=4n(S) = 4. Let AA be the event of getting at least one head. A={HH,HT,TH}  ⟹  n(A)=3A = \{HH, HT, TH\} \implies n(A) = 3. Alternatively, let A′A' be the event of getting no heads. A′={TT}  ⟹  n(A′)=1A' = \{TT\} \implies n(A') = 1. P(A′)=14P(A') = \frac{1}{4} P(A)=1−P(A′)=1−14=34P(A) = 1 - P(A') = 1 - \frac{1}{4} = \frac{3}{4}

Explanation:

The probability of 'at least one' is often easier to calculate using the complement 'none'. Here, the complement of getting at least one head is getting two tails (no heads).