Data Science Wizardry Blog by Attila Vajda

Building paragraphs at the break of dawn

difference between terms #

$k_A$

$g_A$

$P(A)$

\begin{equation} P\left(A_1+A_2+\ldots+A_n\right)=P\left(A_1\right)+P\left(A_2\right)+\ldots+P\left(A_n\right) \end{equation}

The probability of any of the events occurring, is the sum of the probabilities of individual events.

This formula seemed incomprehensible, and intimidating a few days ago, but now I can read it!

What is the likelihood of two dice rolls resulting in a sum of 5? Wow! Each dice have 1/6 probability to result in a given number. Why are the probabilities multiplied? Because 36 outcomes are possible. Out of 36 possibilities, 4 sum to 5, 4/36, or 1/9 is the solution.

I don't understand the following: If A and B are arbitrary events. P(A + B) = P(A) + P(B) - P(A * B)

I struggle with this, became anxious, but I am making effort to understand it. I asked computer agent for analogies.

\begin{equation} P(A+B)=P(A \backslash B)+P(B)=P(A)-P(A \cdot B)+P(B), \end{equation}

Következmény. Ha $A, B$ és $C$ tetszöleges események (5. ábra) akkor $$ \begin{aligned} & P(A \cup B \cup C)= \ = & P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C) . \end{aligned} $$

Ui. legyen $D=B \cup C$, akkor $$ \begin{aligned} A \cap D=A \cap(B \cup C)= & (A \cap B) \cup(A \cap C) \text { és } \ P(A \cap D)=P(A \cap B)+ & P(A \cap C)-P(A \cap B \cap A \cap C)= \ & =P(A \cap B)+P(A \cap C)-P(A \cap B \cap C) . \end{aligned} $$

$$ \begin{aligned} & =P(A)+P(B)+P(C)-P(B \cdot C)-[P(A \cdot B)+P(A \cdot C)-P(A \cdot B \cdot C)]= \ & \quad=P(A)+P(B)+P(C)-P(B \cdot C)-P(A \cdot B)-P(A \cdot C)+P(A \cdot B \cdot C) . \end{aligned} $$

How to make a puzzle of these formulas?