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91Ó°ÊÓ

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

Expert verified
Answer: The unconditional probability of event A occurring is 38%.

Step by step solution

01

Identify the given probabilities

We are given the following probabilities: - Probability of the first subgroup: \(P(S_1) = 0.6\) - Probability of the second subgroup: \(P(S_2) = 0.4\) - Probability of event A occurring in the first subgroup: \(P(A|S_1) = 0.3\) - Probability of event A occurring in the second subgroup: \(P(A|S_2) = 0.5\)
02

Apply the theorem of total probability

We will now use the theorem of total probability to find the unconditional probability of event A, regardless of the subgroup it comes from. The theorem of total probability states: \(P(A) = P(A|S_1) \cdot P(S_1) + P(A|S_2) \cdot P(S_2)\)
03

Calculate the unconditional probability of event A

Using the given probabilities, we can calculate the unconditional probability of event A as follows: \(P(A) = (0.3) \cdot (0.6) + (0.5) \cdot (0.4)\) \(P(A) = 0.18 + 0.2\) \(P(A) = 0.38\) The unconditional probability of event A occurring, regardless of the subgroup it comes from, is \(38 \%\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
Conditional probability refers to the likelihood of an event happening given that another event has already occurred. In our exercise, we are looking into "event A" and how it occurs within each subgroup separately.

To better understand, think about how the probability of rain might change if you know it's a cloudy day. Saying the probability of rain on a cloudy day is higher than on a sunny day is an example of conditional probability at work. Here, the event "cloudy day" affects the probability of rain.

In the exercise, conditional probabilities help us understand how likely event A is to occur within each subgroup:
  • For the first subgroup, the probability of A given the subgroup is one of focus, is denoted by \( P(A|S_1) = 0.3 \).
  • Similarly, for the second subgroup, the probability of A is \( P(A|S_2) = 0.5 \).
These numbers are based on the context of each subgroup, much like how hoping for interruptions based on if you work from home or an office might change your expectation.
Probability Subgroups
Probability subgroups hint at a situation where a larger group is divided into smaller parts that can be analyzed individually.

Imagine taking a large classroom and breaking it into smaller study groups. Just as each group's dynamics might differ, a population broken into subgroups might show different probabilities for events.

In our exercise, the entire population is split into two subgroups:
  • The first subgroup makes up \( 60\% \) of the population.
  • The second subgroup accounts for \( 40\% \) of it.
Each subgroup offers a unique setting to examine independently how often event A occurs. By understanding each subgroup's behavior, we get a clearer picture of the overall population's tendency.
Unconditional Probability
Unconditional probability refers to the likelihood of an event occurring without considering any other conditions or prior events. It gives a general probability, free from specific contexts or constraints.

In our exercise, we are asked to find the unconditional probability of event A. This means we want to know how often event A happens, regardless of which subgroup it arises from.

To find this, we use the theorem of total probability, which combines the weighted probabilities of event A occurring in each subgroup:
  • Multiply the probability of event A in the first subgroup \( P(A|S_1) \) by the probability of being in that subgroup \( P(S_1) \).
  • Do the same for the second subgroup, multiplying \( P(A|S_2) \) by \( P(S_2) \).
  • Add these results together to get the overall probability \( P(A) \).
The calculation was \( P(A) = (0.3 \times 0.6) + (0.5 \times 0.4) = 0.38 \), or an unconditional probability of 38\%. This overall measure helps us understand the general likelihood of event A occurring across the entire population.

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