PROBABILITY IN ARTIFICIAL INTELLIGENCE

REVISED: Wednesday, February 4, 2015

I.  PROBABILITIES

We will start off by using a coin flip with H for heads and T for tails.

Probability of heads is P(H) = 1/2 = 0.5
Probability of tails is P(T) = 1/2 = 0.5

What is probability out of three coin flips of getting three H?

P = { H, H, H } =  ( ( 1/2 * 1/2 ) * 1/2 ) = 1/8 = 0.125

A.  Symbols

The " | " reads "provided that, condition on, or given."
The " : "  reads "such that."
The " ¬ "  reads "not."
The " ⊥ "  reads "independent."
The " ⇒ "  reads "implies."
The " ∑ "  reads "summation."
The " ∩ "  reads "intersection of."

B.  Complimentary Probability

If probability of an event is:

P( A ) = p;

then complimentary probability of an event:

P( ¬A ) is ( 1 - p ).

You can say X and Y are independent events or:

X ⊥ Y

P(A) = p  ⊥  P( ¬A ) is ( 1 - p )

X ⊥ Y : P(X) P(Y) = P(X,Y)

C.  Independence

P = ( X ₁ = H ) = 1/2
H : P ( X ₂  = H  |  X ₁  = H ) = 0.9
T : P ( X ₂  = T  |  X ₁  = T) = 0.8

What is the probability of the second coin flip coming up heads?

P = (X ₂  = H) = 0.55

P = (X ₂  = H) = ((P ( X ₂  = H  |  X ₁  = H )) * (P ( X ₁ = H ))) +
(P ( X ₂  = H  |  X ₁  = T )) * (P ( X ₁ = T )) =
(0.5 * 1/2) + ( ( 1 - 0.8 ) * 1/2) =
(0.45 + 0.1) = 0.55

X _i = Result of i-th coin flip.

X _i  = { H, T }

P(H) = 1/2

P(T) = 1/2

What is probability of all four flips being either H;
or all four flips being T.

Four all four flips being H we have:

P(X ₁ = X ₂ = X ₃ = X ₄ ) =
(( 1/2 * 1/2) * 1/2 ) * 1/2) = 1/16

For all four flips being tails we would have the same thing 1/16.

The probability of either 4 H or 4 T = 1/16 + 1/16 = 2/16 = 1/8.
1/8 = 0.125.

What is probability of at least three out of four flips being three or more H.

P({ X ₁ ,  X ₂ ,  X ₃ ,  X ₄  } contains  >=  3H ) = (5 * 1/16) = 5/16 = 0.3125

HHHH = (((1/2 * 1/2) * 1/2 )* 1/2) = 1/16
HHHT = (((1/2 * 1/2) * 1/2 )* 1/2) = 1/16
HHTH = (((1/2 * 1/2) * 1/2 )* 1/2) = 1/16
HTHH = (((1/2 * 1/2) * 1/2 )* 1/2) = 1/16
THHH = (((1/2 * 1/2) * 1/2 )* 1/2) = 1/16

D. Total Probability 

P( Y ) = S_i P( Y | X = i) P( X = i )

P( ¬X | Y ) = 1 - P( X | Y )

Example 1

P( D ₁ )       P( D ₁  = sunny ) = 0.9
P( D ₂  = sunny | D ₁  = sunny ) = 0.8

P( D ₂  = rainy | D ₁  = sunny ) = 1 - 0.8 = 0.2

Example 1.1

P( D ₂  = sunny | D ₁  = rainy) = 0.6

P( D ₂  = rainy | D ₁  = rainy) = 1 - 0.6 = 0.4 

Example 1.2

P( D ₂  = sunny ) = ( 0.9 * 0.8 )  + ( 1 - 0.9)*0.6) = 0.78

P( D ₃  = sunny ) = ( 0.78 * 0.8) + ( ( 1- 0.78 ) * 0.6) = 0.756

Example 2

P( C ) = 0.01

P( ¬C ) = ( 1.0 - 0.01 ) = 0.99

Example 2.1

P( + | C ) = 0.9

P( - | C ) = ( 1.0 - 0.9 ) = 0.1

P( + | ¬C ) = 0.2

P( - | ¬C ) = ( 1.0 - 0.2 ) = 0.8

Example 2.2

Joint Probabilities

P( +, C ) = ( 0.01 * 0.9 ) = 0.009

P( -, C ) = ( 0.01 * 0.1 ) = 0.001

P( +, ¬C ) = ( 0.99 * 0.2 ) = 0.198

P( -, ¬C ) = ( 0.99 * 0.8 ) = 0.792

Example 2.3

P( C | + ) = ( 0.009 / (0.009 + 0.198 ) ) = ( 0.009 / 0.207 ) = 0.043

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