## Probability

# Axiomatic Approach to Probability

- Probability : Number P(ω
_{i}) associated with sample point ω_{i}such that

(i) 0 ≤ P (ω_{i}) ≤ 1

(ii) Σ P(ω_{i}) for all ω_{i}∈ S = 1

(iii) P(A) = Σ P(ω_{i}) for all ω_{i}∈ A. The number P(ω_{i}) is called probability of the outcome ω_{i} - Equally likely outcomes : All outcomes with equal probability
- Probability of an event: For a finite sample space with equally likely outcomes

Probability of an event P(A) = \frac{n(A)}{n(S)}, where n(A) = number of elements in the set A, n(S) = number of elements in the set S. - If A and B any two events, then

P(A or B) = P(A) + P(B) − P(A and B)

equivalently, P(A ∪ B) = P(A) + P(B) − P(A ∩ B) - If A and B are mutually exclusive, then P(A or B) = P(A) + P(B)
- If A is any event, then

P(not A) = 1 − P(A) - If n letter corresponding to n envelopes are placed in the envelopes at random, then

(i) Probability that all letters are in right envelopes = 1/n!.

(ii) Probability that all letter are not in right envelopes = 1 - \frac{1}{n!}.

(iii) Probability that no letter is in right envelopes = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - .... + (-1)^{n} \frac{1}{n!}

Probability that exactly r letters are in right envelopes = \frac{1}{r!}\left[\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - .... + (-1)^{n-r} \frac{1}{(n-r)!}\right] - If odds in favour of an event are a : b, then the probability of the occurrence of that event is \frac{a}{a + b} and the probability of non-occurrence of that event is \frac{b}{a + b}.
- If odds against an event are a : b, then the probability of the occurrence of that event is \frac{b}{a + b} and the probability of non-occurrence of that event is \frac{a}{a + b}.
- Let A, B, and C are three arbitrary events. Then

(i) Only A occurs is A \cap \overline{B} \cap \overline{C}

(ii) Both A and B, but not C occur is A \cap B \cap \overline{C}

(iii) All the three events occur is A ∩ B ∩ C

(iv) At least one occurs is A ∪ B ∪ C

(v) At least two occur is (A ∩ B) ∪ (B ∩ C) ∪ (A ∩ C)

(vi) None occurs is \overline{A} \cap \overline{B} \cap \overline{C} = \overline{A \cup B \cup C}

(vii) Exactly one of A and B occurs is \left(A \cap \overline{B}\right) \cup \left(\overline{A} \cap B\right) - P (exactly one of them out of three events A, B and C)

\tt =P(A \cap \overline{B} \cap \overline{C}) + P(\overline{A} \cap B \cap \overline{C}) + P(\overline{A} \cap \overline{B} \cap C) - P (exactly two of them out of three events A, B and C)

\tt =P(A \cap B \cap \overline{C}) + P(A \cap \overline{B} \cap C) + P(\overline{A} \cap B \cap C) - \tt P(A \cap \overline{B}) + P(\overline{A} \cap B) = P(A) + P(B) - 2P(A \cap B) = P(A ∪ B) − P(A ∩ B)
- If A and B are any two events, then

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) - If A, B, C are any three events, then P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(C ∩ A) + P(A ∩ B ∩ C)
**Demorgan's laws**

\tt P(\overline{A \cup B}) = P(\overline{A} \cap \overline{B})

\tt P(\overline{A \cap B}) = P(\overline{A} \cup \overline{B})- The probability of odds favarable to an event A is denoted by \tt P(A) : P(\overline{A})
- The probability of odds against to an event A is denoted by \tt P(\overline{A}) : P(A)
- If P(A) is the probability of occurrence of an event A, then the probability of nonoccurrence of an event A is denoted by \tt P(\overline{A}).

\tt \therefore P(A) + P(\overline{A}) = 1 - If A is Null event (or) Impossible event then P(A) = 0
- If A is sure event (or) certain event, then P(A) = 1
- If A and B are mutually exclusive events, then P(A ∩ B) = 0

P(A ∪ B) = P(A) + P(B) - If A and B are exhaustive events, then P(A ∪ B) = 1
- If A, B, C are any three events, then

0 ≤ P(A ∪ B) ≤ 1

0 ≤ P(A ∪ B ∪ C) ≤ 1

P(A) ≤ P(A ∪ B), P(B) ≤ P(A ∪ B)

P(A ∩ B) ≤ P(A), P(A ∩ B) ≤ P(B)

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- (i) P (A ∪ A') = S

(ii) P (A ∩ A') = Φ

(iii) P (A')' = A - If a set of events A
_{1}, A_{2}, ....., A_{n}are mutually exclusive, then A_{1}∩ A_{2 }∩ A_{3}∩..... ∩ A_{n}= Φ

∴ P(A_{1}∪ A_{2 }∪ A_{3}∪ ..... ∪ A_{n}) = P(A_{1}) + (A_{2}) + .... + P(A_{n}) and P(A_{1}∩ A_{2 }∩ A_{3}∩..... ∩ A_{n}) = 0 - If a set of events, A
_{1}, A_{2}, ..... A_{n}are exhaustive, then P(A_{1}∪ A_{2 }∪ ..... ∪ A_{n}) = 1 - Probability of any event in a sample space is 1. i.e., P(A) = 1
- Odds in favour of A=\frac{P(A)}{P(\overline{A})}
- Odds in against of A=\frac{P(\overline{A})}{P(A)}
- For two events A and B

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) - For three events A, B and C

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C) - P(\bar{A} \cap B)=P(B)-P(A \cap B)
- P(A \cap \bar{B})=P(A)-P(A \cap B)
- P[(A \cap \bar{B}) \cup(\bar{A}\cap B)]=P(A)+P(B)-2P(A \cap B)
- P(\bar{A} \cap \bar{B})=1-P(A \cup B)
- P(\bar{A} \cup \bar{B})=1-P(A \cap B)
- P(A) = P(A \cap B)+P(A \cap \bar{B})
- P(B) = P(A \cap B)+P(B \cap \bar{A})
- P(\bar{A}) = 1-P(A)
- P(A\cup \bar{A}) = P(S)=1, P(\phi)=0
- P(A ∩ B) = P(A) × P(B), if A and B are independent events.