# CHSE Odisha Class 11 Math Notes Chapter 16 Probability

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 16 Probability will enable students to study smartly.

## CHSE Odisha 11th Class Math Notes Chapter 16 Probability

Random Or Statistical Experiment:
A random or statistical experiment is one in which

• All possible outcomes of the experiment are known in advance.
• The performance of an experiment result in an outcome is not known in advance.
• The experiment can be repeated under identical conditions.

Sample Space: Sample space is the set of all possible outcomes of an experiment.

Elementary event. An element of sample space is an elementary event.

Event: An event is a subset of a sample space.

Probability of an event: Probability of an event ‘A’ = $$P(A)=\frac{\text { Size of } A}{\text { Size of } S}$$

Types Of Event:

(a) Impossible event
Φ ⊂ S known as the impossible event P(Φ) = 0

(b) Sure (certain) event:
S ⊂ S known as the sure event. P(S) = 1

(c) Mutually exclusive events:
Two events A and B are mutually, exclusive if A ∩ B = Φ i.e occurence of one excludes the occurence of the other.

(d) Equally likely events:
Two events A and B are equally likely if P(A) = P(B).

(e) Independent events:
Two events are independent if occurence if does not depend on occurence of the other.

(f) Exhaustive events:
The events E1, E2, ….. En are exhaustive if E1 ∪ E2 ….. ∪ En = S.

Verbal description of events:
Not a → Ac or $$\overline{\mathrm{A}}$$ or A’
A or B (at least one of A or B) → A ∪ B
A and B → A ∩ B
A but not B → A ∩ Bc
Neither A nor B → Ac ∩ Bc = (A ∪ B)c
Exactly one of A, B or C → (A ∩ Bc ∩ Cc) ∪ (Ac ∩ B ∩ Cc) ∪ (Ac ∩ Bc ∩ Cc).
Exactly two of A, B or C → (A ∩ B ∩ Cc) ∪ (A ∩ Bc ∩ C) ∪ (Ac ∩ B ∩ C)

Some Theorems On Probability:

(a) For any event A: 0 ≤ P(A)’ ≤ 1

(b) P(Φ) = 0, P(S) = 1

(c) P(Ac) = 1 – P(A)

(d) For any two events if A ⊆ B then P(A) ≤ P(B).

(e) For any two events A and B. P(A – B) = P(A ∩ Bc) = P(A) – P(A ∩ B)

(f) For any two events A and B P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

(g) If A and B are mutually exclusive then P(A ∪ B) = P(A) + P(B)

(h) For any three events A, B and C P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)