Question

A population can be divided into two subgroups that occur with probabilities \(60 \%\) and \(40 \%,\) respectively. An event A occurs \(30 \%\) of the time in the first subgroup and \(50 \%\) of the time in the second subgroup. What is the unconditional probability of the event A, regardless of which subgroup it comes from?

Short answer

Answer: The unconditional probability of event A occurring in the given population is 38%.

Step-by-step solution

Identify the probabilities of the subgroups

We are given that the population can be divided into two subgroups with probabilities \(60 \%\) and \(40 \%,\) respectively. We can represent these probabilities as: \(P(S_1) = 0.6\) for the first subgroup, \(P(S_2) = 0.4\) for the second subgroup.

Identify the probabilities of event A occurring within each subgroup

We are also given the following information about the event A occurring within each subgroup: \(P(A|S_1) = 0.3\) for the first subgroup, \(P(A|S_2) = 0.5\) for the second subgroup.

Use the Law of Total Probability to calculate the unconditional probability of event A

According to the Law of Total Probability, the unconditional probability of event A can be calculated as follows: \(P(A) = P(A|S_1) \cdot P(S_1) + P(A|S_2) \cdot P(S_2)\) Now, plug in the given probabilities and solve: \(P(A) = (0.3)(0.6) + (0.5)(0.4) = 0.18 + 0.2 = 0.38\)

Interpret the result

The unconditional probability of event A, regardless of which subgroup it comes from, is \(38 \%\).

Key concepts

Unconditional Probability

Unconditional probability, often termed as marginal probability, is the likelihood of an event happening without any conditions or restrictions. It represents the overall probability of an event occurring, taking into account all possible scenarios that could lead to it.
This concept is crucial because it helps us understand the probability of an event in a complete system, without focusing on a specific condition. In the original exercise, we calculated the unconditional probability of event A without specifying which subgroup it belongs to.
The unconditional probability of event A was found using the Law of Total Probability. This involves summing up the probabilities of event A occurring within each subgroup, weighted by the probability of each subgroup. The formula used is:

• Given two subgroups, calculate:
  \[ P(A) = P(A|S_1) \cdot P(S_1) + P(A|S_2) \cdot P(S_2) \]
• For this particular problem, it resulted in:\[ P(A) = (0.3)(0.6) + (0.5)(0.4) = 0.38 \]

This shows that event A has a 38% chance of occurring irrespective of which subgroup it arises from.

Conditional Probability

Conditional probability is the likelihood of an event happening given that another event has already occurred. It's denoted as \( P(A|B) \), which reads "the probability of A given B." This concept allows us to zero in on the probability of event A occurring, assuming event B is true.
In the context of the original exercise, conditional probability played a central role. We used the probabilities of event A occurring within each subgroup, written as:

• \( P(A|S_1) = 0.3 \)
• \( P(A|S_2) = 0.5 \)

These conditional probabilities inform us of the distinct likelihood of event A in each separate subgroup setup or condition. Understanding conditional probabilities enables more nuanced analysis, especially in cases where different segments or conditions affect the probabilities of outcomes.

Probability Distribution

Probability distribution provides a complete picture of how probabilities are allocated across different outcomes or scenarios in a system. It encapsulates all possible events and their likelihoods, ensuring that the sum equals one. Probability distribution is the foundation that helps evaluate both unconditional and conditional probabilities by organizing and structuring all relevant probabilities within a specific context or population.
In our exercise, the probability distribution was implicit in two dimensions: the division of the population into subgroups and the occurrence of event A within these subgroups.
This can be seen through:

• The distribution of the population across subgroups:\( P(S_1) = 0.6 \) and \( P(S_2) = 0.4 \)
• Within each subgroup, how event A's occurrence is distributed: \( P(A|S_1) = 0.3 \) and \( P(A|S_2) = 0.5 \)

These probabilities add up to describe how likely each scenario is, thereby giving us a complete framework to work with probability calculations like those in the solution. With probability distribution, we can deal with complex situations in a structured and meaningful way.