Calculation of the probability of an emergency. OBZhD

BASICS OF SAFE LIFE
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Calculation of the probability of an emergency. Basics of safe life

Fundamentals of Safe Life Activities (OBZhD)

Emergencies create damage that may or may not be quantifiable (eg deaths, personal injury, property damage, environmental damage, etc.). For the purpose of unification, various consequences and harm are denoted by the term "damage". Damage is measured in terms of money or the number of deaths, the number of injured people, etc. To measure damage in value terms, it is necessary to find an equivalent between these units of measurement.

Calculation of emergency probabilities (state of emergency). Let P{E} denote its probability. Probability of a certain event P{E} = 1, probability of an impossible event P{E} = 0, probability of the sum of pairwise incompatible PEs (E_j E_j is not zero if i ≠ j) is

PE E_i , E_j ..., E_n , form a complete group of events if they are pairwise incompatible and one of them necessarily occurs for the complete group of events

In particular, for equally probable CPs (P{E} = p,i =1, 2, ..., n) forming a complete group of events, the probability of an emergency

P = 1/n.

Opposite events E and E form a complete group, so

A complete group of events can be identified using the Karnot map. Three emergencies X, Y, Z form a Karnot map. The PEs written in the cells are pairwise incompatible.

When the number of emergencies exceeds five, Karnot cards are inconvenient to use. Then the complete group of events can be generated using binary numbers. For n emergencies write decimal numbers from 0 to (2ⁿ - 1) and their representations in the binary system.

Let us determine the probability (P) of the emergency. R-NP is the sum a and N. Accident N and accident А can come together. Therefore, the formula for determining the probability of pairwise incompatible events P{S} is unsuitable. Selecting a complete group of events using the Karnaugh map, we find the probability of P‑CHP:

P{A + N} = P{A} + P{N} - Р {AN}.

If catastrophe (K) is impossible, К = AN is not equal to zero, then P{AN} = 0.

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