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

The outcomes satisfying a particular condition are called favourable outcomes.

A set of favourable outcomes of a given sample space is an ‘event’. Event is a subset of the sample space.

• Events are denoted by capital letters A, B, C, D etc.
• The number of elements in event A is written n(A).
• Since A ⊆ S, always 0 ≤ n(A) ≤ n(S).
• Example: two coins tossed, A = getting at least one tail → A = {TT, TH, HT}, n(A) = 3.

If A and B are two events associated with the same sample space of a random experiment, the conditional probability of the event A given that B has occurred, i.e. P(A∣B), is given by

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: \[P(A' | B) = 1 - P(A | B)\], P(B) ≠ 0

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:

• \[P(E \cap F) = P(F) \cdot P(E | F)\],\[ P(F) \neq 0. \]

• \[P(E \cap F) = P(E) \cdot P(F | E)\],\[ P(E) \neq 0. \]

For three events:

If B₁, B₂,..., B_n 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,

• The sample space \[ S \] contains all possible outcomes.
• An event is a subset of the sample space.
• A simple event has one outcome; a compound event has more than one outcome.
• A certain event has probability \[ 1; \] an impossible event has probability \[ 0. \]
• The complement of \[ A \] is \[ A', \] and \[ P(A') = 1 - P(A). \]
• Mutually exclusive events cannot occur together: \[ A \cap B = \varnothing. \]
• Exhaustive events together cover the entire sample space: \[ A \cup B = S. \]
• For equally likely outcomes, \[ P(A) = \frac{n(A)}{n(S)}. \]
• For any two events, \[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \]

• Conditional probability means probability under a given condition.

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

• First restrict the sample space according to the given condition, then count the favourable outcomes.

• 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), provided P(B) ≠ 0.

• 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)