Revision: Probability Maths HSC Science (General) 11th Standard Maharashtra State Board

Probability measures the degree of certainty of the occurrence of an event.

The conditional probability of both events A and B over the sample space S is

Two events are said to be independent if the occurrence of one does not depend on the other.

For two events E and F:

• E and F are independent if P(F | E) = P(F), when \[P(E) \neq 0\].

• Similarly, E and F are independent if P(E | F) = P(E), when \[P(F) \neq 0\].

• An equivalent and most commonly used test is:

\[P(A | B) = \frac{P(A \cap B)}{P(B)}\], where \[P(B) \neq 0\].

\[P(B | A) = \frac{P(A \cap B)}{P(A)}\], where \[P(A) \neq 0\].

Multiplication rule: \[P(A \cap B) = P(B) \cdot P(A | B) = P(A) \cdot P(B | A)\].

Complement form: \[P(A' | B) = 1 - P(A | B)\].

If A and B are any two events defined over a sample space S, then P(A∪B) = P(A) + P(B) – P(A∩В)

or

P (A+B) = P(A) + P(B) – P(AB)

Mutually Exclusive Events:

If P(A ∩ B) = 0, then

P(A ∪ B) = P(A) + P(B)

For two events:

• \[P(E \cap F) = P(F) \cdot P(E | F)\]

• \[P(E \cap F) = P(E) \cdot P(F | E)\]

For three events:

If B₁, B₂,..., Bn are mutually exclusive and exhaustive events and if A is an event consequent to these B_i's, then for each i = 1, 2, 3, ..., n,

Properties:

• Complement Rule
  P(A′) = 1 − P(A)
  ⇒ P(A) + P(A′) = 1
• Range of Probability
  0 ≤ P(A) ≤ 1
• Impossible Event
  P(ϕ) = 0
• Certain Event
  P(S) = 1
• Subset Rule
  If A ⊆ B, then P(A) ≤ P(B)
• Difference of Events
  P(A ∩ B′) = P(A) − P(A ∩ B)
  P(A′ ∩ B) = P(B) − P(A ∩ B)
• Union of Two Events
  P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
• Union of 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)
• Mutually Exclusive Events (2 events)
  If A ∩ B = 0, then
  P(A ∪ B) = P(A) + P(B)
• Mutually Exclusive Events (multiple)
  P(A₁ ∪ A₂ ∪ ... ∪ Aₙ) = P(A₁) + P(A₂) + ... + P(Aₙ)
• Upper Bound of Union
  P(A ∪ B) ≤ P(A) + P(B)

Playing Cards – Key Facts

• Total cards = 52

• Red cards = 26 (Hearts, Diamonds)

• Black cards = 26 (Clubs, Spades)

• Each suit has 13 cards

• Face cards = King, Queen, Jack (Total = 12)

• Conditional probability means probability under a given condition.

• The formula is \[P(A | B) = \frac{P(A \cap B)}{P(B)}\], where \[P(B) \neq 0\].

• Always reduce the sample space according to the condition first.

• The numerator represents outcomes common to both events.

• Do not confuse P(A | B) with P(B | A).

• For independent events, P(A | B) = P(A).

• Multiplication theorem is used to find the probability of simultaneous occurrence of events.

• For two events: \[P(E \cap F) = P(E) \cdot P(F | E)\]

• Another equivalent form is \[P(E \cap F) = P(F) \cdot P(E | F)\].

• For three events: \[P(E \cap F \cap G) = P(E) \cdot P(F | E) \cdot P(G | E \cap F)\].

• Most “without replacement” questions are solved using this theorem.

• Always define events before solving a probability problem.

• Independent events do not influence each other.

• The standard test is \[P(E \cap F) = P(E)P(F)\].

• Conditional form: \[P(F | E) = P(F)\] and \[P(E | F) = P(E)\], when defined.

• If two events are independent, related complement pairs are also independent.

• Mutually exclusive events and independent events are different.

• For independent events A and B, \[P(A \cup B) = 1 - P(A')P(B')\].

• For three events, mutual independence requires pairwise conditions and the condition involving all three together.

• Bayes' Theorem works from effect to cause.

• Prior probability means “before observation”.

• Posterior probability means “after observation”.

• The observed event is usually given in the question statement.

• The required answer is often a probability of the form P(cause ∣ observed event).

1. Odds in favour of event A
   Odds in favour of A = P(A) / P(A′)
2. Odds against event A
   Odds against A = P(A′) / P(A)