\mathbb{P}: \mathcal{F} \rightarrow [0,1]

X: \mathcal{\Omega} \rightarrow \mathbb{R}

X_1, X_2, \dots, X_t,\dots

(ML)

P(A) = ?

P(A) \geq 0~\forall A

P(\Omega) = 1

A_1, A_2, \dots, A_n

= \sum_{i=1}^n P(A_i)

P(A_1\cup A_2 \cup \dots \cup A_n)

= \sum_{i=1}^n P(A_i)

P(A_1\cup A_2 \cup \dots \cup A_n)

A_1

A_2

A_3

A_4

A_5

A_6

P(A_1),P(A_2),P(A_3),P(A_4),P(A_5),P(A_6)

B

C

P(B) = P(A_1)+P(A_3)+P(A_5)

\geq 5

P(C) = P(A_5)+P(A_6)

D

P(D) = P(A_3)+P(A_6)

P(A) = 1 - P(A^\mathsf{c})

A \cup A^\mathsf{c} = \Omega

1 = P(\Omega) = P(A \cup A^\mathsf{c}) = P(A) + P(A^\mathsf{c})

\therefore P(A) = 1 - P(A^\mathsf{c})

P(A) \leq 1

P(A) = 1 - P(A^\mathsf{c})

P(A \cup B) = P(A) + P(B) - P(A \cap B)

A^\mathsf{c}

P(A \cup B) = P(A \cup (B \cap A^\mathsf{c}))

= P(A) + P(B \cap A^\mathsf{c})

P(B) = P((B \cap A^\mathsf{c}) \cup (B \cap A))

= P(B\cap A^\mathsf{c}) + P(B\cap A)

\therefore P(B\cap A^\mathsf{c}) = P(B) - P(B\cap A)

= P(A) + P(B) - P(B\cap A)

\Omega = A_1 \cup A_2 \cup \dots \cup A_n

P(\Omega) = P(A_1 \cup A_2 \cup \dots \cup A_n) = \sum_{i=1}^n P(A_i)

\therefore \sum_{i=1}^n P(A_i) = 1

\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}

\Omega

P(\phi) = 0

\therefore 1 = P(\Omega) = P(\Omega \cup \phi) = P(\Omega) + P(\phi)

\therefore 1 = 1 + P(\phi) \implies P(\phi) = 0

A_0, A_1, A_2, A_3, A_4, A_5, A_6

P(A_0)=0.3, P(A_1)=0.45, P(A_2)=0.12, P(A_3)=0.02, \\ P(A_4)=0.07, P(A_5)=0.01, P(A_6)=0.03

X = A_0 \cup A_2 \cup A_4 \cup A_6

P(X) = P(A_0 \cup A_2 \cup A_4 \cup A_6)

= P(A_0) + P(A_2) + P(A_4) + P(A_6)

A_0, A_1, A_2, A_3, A_4, A_5, A_6

P(A_0)=0.3, P(A_1)=0.45, P(A_2)=0.12, P(A_3)=0.02, \\ P(A_4)=0.07, P(A_5)=0.01, P(A_6)=0.03

X = A_0 \cup A_1 \cup A_2 \cup A_3 \cup A_4

P(X) = P(A_0 \cup A_1 \cup A_2 \cup A_3 \cup A_4)

= P(A_0) + P(A_1) + P(A_2) + P(A_3) + P(A_4)

A_0, A_1, A_2, A_3, A_4, A_5, A_6

P(A_0)=0.3, P(A_1)=0.45, P(A_2)=0.12, P(A_3)=0.02, \\ P(A_4)=0.07, P(A_5)=0.01, P(A_6)=0.03

X_1 = \{0, 2, 4, 6\} = A_0 \cup A_2 \cup A_4 \cup A_6

X_2 = \{0, 3, 6\} = A_0 \cup A_3 \cup A_6

X_1 \cap X_2 = \{0, 6\} = A_0 \cup A_6

P(X_1) = 0.3 + 0.12 + 0.07 + 0.03 = 0.52

P(X_2) = 0.3 + 0.02 + 0.03 = 0.35

P(X_1 \cap X_2) = 0.3 + 0.03 = 0.33

P(X_1 \cup X_2) = P(X_1) + P(X_2) - P(X_1 \cap X_2)

\therefore P (X_1 \cup X_2) = 0.52 + 0.35 - 0.33 = 0.54

P(A_1)=0.01, P(A_2)=0.03, P(A_1 \cap A_2) = 0.005

P(A_1 \cup A_2)= P(A_1) + P(A_2) - P(A_1 \cap A_2)

= 0.01 + 0.02 - 0.005 = 0.025

A_2

A_1

P(A_1)=0.01, P(A_2)=0.03, P(A_1 \cap A_2) = 0.005

P(A_1 \cup A_2)^\mathsf{c}= 1 - P(A_1 \cup A_2)

= 1 - 0.025 = 0.975

A_2

A_1

P(H) = \frac{12012}{24000} = 0.5005

P(A_i) = \frac{no.~of~times~the~outcome~is~in~A_i}{total~no.~of~times~the~experiment~was~repeated}

P(A_i) \geq 0~?

P(\Omega)=1~?

P(\Omega) = \frac{no.~of~times~the~outcome~is~in~S}{total~no.~of~times~the~experiment~was~repeated} = 1

\Omega

P(A_1 \cup A_2) = P(A_1)+ P(A_2)~?

P(A_1 \cup A_2) = \frac{k_1 + k_2}{k} = \frac{k_1}{k} + \frac{k_2}{k}

= P(A_1)+ P(A_2)

A_1

A_2

\Omega

P(A_1 \cup A_2) = P(A_1)+ P(A_2)~?

|\Omega| = n \implies 2^n~subsets \implies 2^n~events

A_1, A_2, A_3, \dots, A_n

A_1, A_2, A_3, \dots, A_n

P(forest) = \frac{15000}{100000} = 0.15

P(infected) = \frac{1000000}{20000000} = 0.05

Flu

\Omega

P(H) = P(T) = k

\Omega = H \cup T

P(\Omega) = P(H \cup T)

= P(H) + P(T)

\Omega = \{H, T\}

H

T

= 2k

= 1

\therefore P(H) = P(T) = k = \frac{1}{2}

\Omega

\phi, \{H\}, \{T\}, \{H, T\}

A_i

A_1, A_2, A_3, A_4, A_5, A_6

\Omega

P(A_1) = P(A_2) = P(A_3) = P(A_4) = P(A_5) = P(A_6) = k

P(S) = \sum_{i=1}^6 P(A_i) = 6k = 1

\therefore P(A_i) = \frac{1}{6}

\Omega

\Omega = \{1, 2, 3, 4, 5, 6\}

A_1

A_2

A_3

A_4

A_5

A_6

\Omega

E

\Omega = \{1, 2, 3, 4, 5, 6\}

A_1

A_2

A_3

A_4

A_5

A_6

O

D

E

P(E) = \sum_{i=1}^{k} \frac{1}{n} = \frac{k}{n}

\frac{1}{n}

P(X) = \frac{number~of~outcomes~in~X}{number~of~outcomes~in~\Omega}

\frac{1}{n}

P(X) = \frac{number~of~outcomes~in~X}{number~of~outcomes~in~\Omega}

P(A_i) \geq 0~?

P(\Omega)=1~?

P(A_1 \cup A_2) = P(A_1)+ P(A_2)~?

A_1

A_2

P(A_1 \cup A_2) = \frac{k_1 + k_2}{n} = \frac{k_1}{n} + \frac{k_2}{n}

= P(A_1) + P(A_2)

\frac{1}{n}

P(B) = \frac{26}{52}

P(X) = \frac{number~of~outcomes~in~X}{number~of~outcomes~in~\Omega}

n = {52 \choose 3} = 22100

P(A) = \frac{4}{22100}

P(C) = \frac{\pi r^2}{\pi R^2} = (\frac{r}{R})^2

R

r

\Omega = \{1,2,3,\dots\}

P(n) = \frac{1}{2^n}

P(n) \geq 0

P(\Omega) = \sum_{i=1}^{\infty}\frac{1}{2^n} = \frac{1}{2}\sum_{i=0}^{\infty} \frac{1}{2^i} = \frac{1}{2}\cdot\frac{1}{1 - \frac{1}{2}} = 1

P(outcome~is~even) = P(\{2\} \cup \{4\} \cup \{6\} \cup \dots )

P(outcome~is~even) = P(2) + P(4) + P(6) + \dots

=\frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} + \dots = \frac{1}{2^2}(1 + \frac{1}{4} + \frac{1}{4^2} + \dots) = \frac{1}{3}

= \sum_{i=1}^\infty P(A_i)

P(A_1\cup A_2 \cup \dots )