Question

If an experiment is conducted, one and only one of three mutually exclusive events \(S_{1}, S_{2},\) and \(S_{3}\) can occur, with these probabilities: $$P\left(S_{1}\right)=.2 \quad P\left(S_{2}\right)=.5 \quad P\left(S_{3}\right)=.3$$ The probabilities of a fourth event \(A\) occurring, given that event \(S_{1}, S_{2},\) or \(S_{3}\) occurs, are $$P\left(A \mid S_{1}\right)=.2 \quad P\left(A \mid S_{2}\right)=.1 \quad P\left(A \mid S_{3}\right)=.3$$ If event \(A\) is observed, find \(P\left(S_{1} \mid A\right), P\left(S_{2} \mid A\right),\) and \(P\left(S_{3} \mid A\right)\).

Short answer

Question: Given the probabilities $P(S_1) = 0.2, P(S_2) = 0.5, P(S_3) = 0.3$, and $P(A|S_1) = 0.2, P(A|S_2) = 0.1, P(A|S_3) = 0.3$, find the conditional probabilities $P(S_1|A)$, $P(S_2|A)$, and $P(S_3|A)$. Answer: The conditional probabilities are $P(S_1|A) = 0.25$, $P(S_2|A) = 0.3125$, and $P(S_3|A) = 0.5625$.

Step-by-step solution

Recall Bayes' theorem

Bayes' theorem states that for events \(B\) and \(C\): $$P(B|C) = \frac{P(C|B) P(B)}{P(C)}$$ In this case, we need to find \(P(S_{1} \mid A)\), \(P(S_{2} \mid A)\), and \(P(S_{3} \mid A)\), where \(B\) is one of the events \(S_1, S_2, S_3\) and \(C\) is the event \(A\).

Express the probabilities

From the given information, we know: $$P(S_{1}) = 0.2, P(S_{2}) = 0.5, P(S_{3}) = 0.3$$ $$P(A|S_{1}) = 0.2, P(A|S_{2}) = 0.1, P(A|S_{3}) = 0.3$$

Calculate the probability of A

We need to compute \(P(A)\), which can be obtained using the law of total probability: $$P(A) = P(A|S_{1})P(S_{1}) + P(A|S_{2})P(S_{2}) + P(A|S_{3})P(S_{3})$$ Plugging the values we know: $$P(A) = 0.2 \cdot 0.2 + 0.1 \cdot 0.5 + 0.3 \cdot 0.3 = 0.16$$

Calculate the conditional probabilities

Now we can use Bayes' theorem to find the probabilities we need: 1. For \(P(S_{1} \mid A)\): $$P(S_{1}|A) = \frac{P(A|S_{1})P(S_{1})}{P(A)} = \frac{0.2 \cdot 0.2}{0.16} = \frac{1}{4} = 0.25$$ 2. For \(P(S_{2} \mid A)\): $$P(S_{2}|A) = \frac{P(A|S_{2})P(S_{2})}{P(A)} = \frac{0.1 \cdot 0.5}{0.16} = \frac{5}{16} = 0.3125$$ 3. For \(P(S_{3} \mid A)\): $$P(S_{3}|A) = \frac{P(A|S_{3})P(S_{3})}{P(A)} = \frac{0.3 \cdot 0.3}{0.16} = \frac{9}{16} = 0.5625$$ So we have found: $$P(S_{1} \mid A) = 0.25, P(S_{2} \mid A) = 0.3125, P(S_{3} \mid A) = 0.5625$$

Key concepts

Conditional Probability

Conditional probability is a fundamental concept in probability theory. It refers to the probability of an event occurring given that another event has already happened. In the context of the given exercise, we are interested in the probabilities of events \(S_1\), \(S_2\), and \(S_3\) happening given that event \(A\) has occurred. This is denoted as \(P(S_i | A)\).

To calculate these probabilities, we use Bayes' Theorem, which helps in updating our probability estimates based on new information. Using Bayes' Theorem, the conditional probability \(P(S_i | A)\) is calculated using the formula:

• \(P(S_i | A) = \frac{P(A|S_i) P(S_i)}{P(A)}\)

Here, \(P(A | S_i)\) is the probability of \(A\) occurring given \(S_i\), \(P(S_i)\) is the initial probability of \(S_i\), and \(P(A)\) is the probability of \(A\) occurring at all. This approach helps to refine probability assessments when additional conditions are known.

Law of Total Probability

The Law of Total Probability is a crucial principle used to determine the probability of an event by considering all possible ways the event can occur. This law states that if \(S_1, S_2,\) and \(S_3\) are exhaustive and mutually exclusive events, the probability of any event \(A\) can be calculated by considering the conditional probabilities with these events.

The formula is:

• \(P(A) = P(A|S_{1})P(S_{1}) + P(A|S_{2})P(S_{2}) + P(A|S_{3})P(S_{3})\)

This formula breaks down the probability of \(A\) into parts based on \(A\)'s occurrence in connection with each \(S_i\).

In the exercise, it helps us calculate \(P(A)\) by adding up the weighted probabilities of \(A\) under different scenarios, where the weights are the probabilities of \(S_i\). Knowing \(P(A)\) is essential for applying Bayes' Theorem, as it serves as the denominator when calculating any conditional probability \(P(S_i|A)\).

Mutually Exclusive Events

Mutually exclusive events are events that cannot happen at the same time. In probability terms, if \(S_1, S_2, \text{and } S_3\) are mutually exclusive, the occurrence of one prevents any others from occurring simultaneously. This means \(P(S_i \cap S_j) = 0\) for any distinct events \(S_i\) and \(S_j\).

In the exercise, \(S_1, S_2,\) and \(S_3\) are mutually exclusive events, which is why the sum of their probabilities equals 1:

• \(P(S_1) + P(S_2) + P(S_3) = 1\)

This property is used to ensure that one and only one of these events will occur, simplifying the calculation of \(P(A)\) using the Law of Total Probability. Understanding mutually exclusive events is essential because it narrows down possible outcomes to non-overlapping probabilities, allowing for straightforward computations when analyzing probability scenarios such as this one.