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Probability - Exhaustive events

Grade 11CBSE

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

🔑Concepts

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Exhaustive events are a set of events in a sample space SS such that at least one of them must occur whenever the experiment is performed.

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Mathematically, a collection of events E1,E2,E3,…,EnE_1, E_2, E_3, \dots, E_n is said to be exhaustive if their union is equal to the sample space: E1∪E2∪E3∪⋯∪En=SE_1 \cup E_2 \cup E_3 \cup \dots \cup E_n = S.

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For exhaustive events, the probability of their union is always 11, i.e., P(E1∪E2∪⋯∪En)=P(S)=1P(E_1 \cup E_2 \cup \dots \cup E_n) = P(S) = 1.

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Events can be exhaustive without being mutually exclusive. If events are both mutually exclusive and exhaustive, they form a 'partition' of the sample space.

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In a partition of sample space SS into nn events E1,E2,…,EnE_1, E_2, \dots, E_n, the sum of their individual probabilities is exactly 11 because they do not overlap and cover the entire space: ∑i=1nP(Ei)=1\sum_{i=1}^{n} P(E_i) = 1.

📐Formulae

E1∪E2∪E3∪⋯∪En=SE_1 \cup E_2 \cup E_3 \cup \dots \cup E_n = S

P(E1∪E2∪E3∪⋯∪En)=1P(E_1 \cup E_2 \cup E_3 \cup \dots \cup E_n) = 1

P(E)+P(E′)=1 (where E and E′ are exhaustive and mutually exclusive)P(E) + P(E') = 1 \text{ (where } E \text{ and } E' \text{ are exhaustive and mutually exclusive)}

💡Examples

Problem 1:

In the experiment of rolling a fair die, let A={1,2,3,4}A = \{1, 2, 3, 4\} and B={3,4,5,6}B = \{3, 4, 5, 6\}. Determine if events AA and BB are exhaustive.

Solution:

The sample space for rolling a die is S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}. To check if AA and BB are exhaustive, find their union: A∪B={1,2,3,4}∪{3,4,5,6}={1,2,3,4,5,6}A \cup B = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} = \{1, 2, 3, 4, 5, 6\}. Since A∪B=SA \cup B = S, the events AA and BB are exhaustive.

Explanation:

Events are exhaustive if their union covers every possible outcome in the sample space. Even though AA and BB share outcomes {3,4}\{3, 4\} (meaning they are not mutually exclusive), they are still exhaustive.

Problem 2:

Consider tossing two coins. Let E1E_1 be the event 'at least one head' and E2E_2 be the event 'at least one tail'. Are E1E_1 and E2E_2 exhaustive?

Solution:

The sample space is S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}. E1={HH,HT,TH}E_1 = \{HH, HT, TH\} E2={HT,TH,TT}E_2 = \{HT, TH, TT\} Finding the union: E1∪E2={HH,HT,TH,TT}=SE_1 \cup E_2 = \{HH, HT, TH, TT\} = S. Since the union equals SS, E1E_1 and E2E_2 are exhaustive events.

Explanation:

By listing all outcomes in the union, we see that every element of SS is contained in either E1E_1 or E2E_2, satisfying the definition of exhaustive events.

Problem 3:

If AA, BB, and CC are three mutually exclusive and exhaustive events associated with a random experiment, and P(A)=0.3P(A) = 0.3, P(B)=0.4P(B) = 0.4, find P(C)P(C).

Solution:

Since A,B,A, B, and CC are mutually exclusive and exhaustive: P(A)+P(B)+P(C)=1P(A) + P(B) + P(C) = 1 Substitute the given values: 0.3+0.4+P(C)=10.3 + 0.4 + P(C) = 1 0.7+P(C)=10.7 + P(C) = 1 P(C)=1−0.7=0.3P(C) = 1 - 0.7 = 0.3

Explanation:

For events that are both mutually exclusive (no overlap) and exhaustive (total coverage), the sum of their probabilities must equal 11.