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Law of Total Probability

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Law of Total Probability#

The Law of Total Probability allows us to compute the probability of an event based on a set of mutually exclusive and exhaustive scenarios.

If \(B_1, B_2, \dots, B_n\) are mutually exclusive and exhaustive events (they cover all possibilities), then for any event \(A\):

\[ P(A) = \sum_{i=1}^{n} P(A \mid B_i) \, P(B_i) \]

Figure: Law of Total Probability; illustrating how an event $A$ can occur under different scenarios $B_1, B_2, B_3$ and how their contributions combine to form the total probability of $A$. Source: WallStreetMojo

Explanation#

  1. Mutually exclusive: No two \(B_i\) events happen at the same time.

  2. Exhaustive: One of the \(B_i\) events must happen.

  3. Conditional probability: \(P(A \mid B_i)\) is the probability of \(A\) given \(B_i\) occurs.

  4. The law sums over all possible scenarios to get the total probability of \(A\).

Mutually exclusive and exhaustive event: Events that cannot happen at the same time (mutually exclusive) and cover all possible outcomes (exhaustive).

Example#

Suppose there are two factories producing light bulbs:

  • Factory 1 produces 60% of all bulbs, with a 1% defect rate.

  • Factory 2 produces 40% of all bulbs, with a 2% defect rate.

Question: What is the probability that a randomly chosen bulb is defective?

Solution using the Law of Total Probability:

\[ P(\text{Defective}) = P(\text{Defective} \mid F_1) P(F_1) + P(\text{Defective} \mid F_2) P(F_2) \]
\[ P(\text{Defective}) = 0.01 \times 0.6 + 0.02 \times 0.4 = 0.014 \]

So, there is a 1.4% chance that a randomly chosen bulb is defective.

### Python Code Example

# Probabilities
P_F1 = 0.6
P_F2 = 0.4
P_def_given_F1 = 0.01
P_def_given_F2 = 0.02

# Law of Total Probability
P_defective = P_def_given_F1 * P_F1 + P_def_given_F2 * P_F2
print("Probability of a defective bulb:", P_defective)

Probability of a defective bulb: 0.014
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