Definitions [4]
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
If a random variable X takes values x₁, x₂, …, xₙ with respective probabilities p₁, p₂, …, pₙ, then it is called the probability distribution of X.
Two events are said to be independent if the occurrence of one does not depend on the other.
For two events E and F:
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E and F are independent if P(F | E) = P(F), when \[P(E) \neq 0\].
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Similarly, E and F are independent if P(E | F) = P(E), when \[P(F) \neq 0\].
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An equivalent and most commonly used test is:
The probability distribution of the number of successes in an experiment consisting of n-Bernoulli trials obtained by the binomial expansion of (q + p )ⁿ is called the binomial distribution.
where p = probability of success and
q = probability of failure
\[P\left(X=r\right)=^{n}C_{r}p^{r}q^{n-r}\] is called probability function.
Formulae [1]
\[P(A | B) = \frac{P(A \cap B)}{P(B)}\], 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
Theorems and Laws [2]
For two events:
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\[P(E \cap F) = P(F) \cdot P(E | F)\],\[ P(F) \neq 0. \]
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\[P(E \cap F) = P(E) \cdot P(F | E)\],\[ P(E) \neq 0. \]
For three events:
- \[P(E \cap F \cap G) = P(E) \cdot P(F | E) \cdot P(G | E \cap F)\], \[ P(E) \neq 0. \] and \[ P(E∩F) \neq 0. \]
If B1, B2,..., Bn are mutually exclusive and exhaustive events and if A is an event consequent to these Bi's, then for each i = 1, 2, 3, ..., n,
Key Points
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Conditional probability means probability under a given condition.
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The formula is \[P(A | B) = \frac{P(A \cap B)}{P(B)}\], where \[P(B) \neq 0\].
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First restrict the sample space according to the given condition, then count the favourable outcomes.
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The numerator represents outcomes common to both events.
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Do not confuse P(A | B) with P(B | A).
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For independent events, P(A | B) = P(A), provided P(B) ≠ 0.
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Multiplication theorem is used to find the probability of simultaneous occurrence of events.
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For two events: \[P(E \cap F) = P(E) \cdot P(F | E)\]
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Another equivalent form is \[P(E \cap F) = P(F) \cdot P(E | F)\].
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For three events: \[P(E \cap F \cap G) = P(E) \cdot P(F | E) \cdot P(G | E \cap F)\].
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Most “without replacement” questions are solved using this theorem.
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Always define events before solving a probability problem.
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Bayes' Theorem works from effect to cause.
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Prior probability means “before observation”.
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Posterior probability means “after observation”.
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The observed event is usually given in the question statement.
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The required answer is often a probability of the form P(cause ∣ observed event).
- Grouped data uses class intervals.
- Class mark \[ x_{i} = \frac{\text{lower limit} + \text{upper limit}}{2}. \]
- Direct method: \[ \bar{x} = \frac{\sum f_{i}x_{i}}{N}. \]
- Assumed mean method: \[ \bar{x} = A + \frac{\sum f_{i}d_{i}}{N}. \]
- Step-deviation method: \[ \bar{x} = A + \left(\frac{\sum f_{i}u_{i}}{N}\right)g. \]
- \[ d_{i} = x_{i} - A, \; u_{i} = \frac{d_{i}}{g}, \; N = \sum f_{i}. \]
- The mean of grouped data is an estimate.
- Always show the calculation table and mention units in board answers.
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Independent events do not influence each other.
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The standard test is \[P(E \cap F) = P(E)P(F)\].
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Conditional form: \[P(F | E) = P(F)\] and \[P(E | F) = P(E)\], when defined.
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If two events are independent, related complement pairs are also independent.
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Mutually exclusive events and independent events are different.
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For independent events A and B, \[P(A \cup B) = 1 - P(A')P(B')\].
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For three events, mutual independence requires pairwise conditions and the condition involving all three together.
Concepts [9]
- Concept of Probability
- Conditional Probability
- Multiplication Theorem of Probability
- Bayes’ Theorem
- Probability Distribution of Discrete Random Variables
- Mean of Grouped Data
- Variance of a Random Variable
- Independent Events
- Probability using Binomial Distribution
