Problem 32 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Given that \(P(A)=1 / 3, P(B)=1 / 4, P(A \mid B)=\) \(1 / 6\), then \(P(B \mid A)\) equal to (a) \(1 / 4\) (b) \(1 / 8\) (c) \(3 / 4\) (d) \(1 / 2\)

Step-by-Step Solution

Answer

The answer is (b) \(1/8\).

Step 1: Understanding the Formula

We need to find the conditional probability \(P(B \mid A)\). Using the formula for conditional probability, we have: \[P(B \mid A) = \frac{P(A \cap B)}{P(A)}.\] Our task now is to find \(P(A \cap B)\) to use in the formula.

Step 2: Use Known Conditional Probability

The problem gives us \(P(A \mid B) = \frac{1}{6}\). Using the formula for conditional probability, \[P(A \mid B) = \frac{P(A \cap B)}{P(B)},\] we can rearrange to find \(P(A \cap B)\): \[P(A \cap B) = P(A \mid B) \cdot P(B).\] Substitute the values: \[P(A \cap B) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}.\]

Step 3: Apply to Original Formula

Now that we have \(P(A \cap B) = \frac{1}{24}\), we can substitute it back into the formula for \(P(B \mid A)\): \[P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{24}}{\frac{1}{3}}.\]

Step 4: Simplify the Expression

To simplify \(\frac{\frac{1}{24}}{\frac{1}{3}}\), we can multiply by the reciprocal: \[P(B \mid A) = \frac{1}{24} \cdot \frac{3}{1} = \frac{3}{24} = \frac{1}{8}.\] Thus, \(P(B \mid A)\) equals \(\frac{1}{8}\).

Key Concepts

Probability of Events

Probability is a way to measure how likely it is for an event to occur. When dealing with **probability of events**, especially in real-world applications, it refers to understanding the likelihood of a given outcome or set of outcomes.
Probability is usually quantified as a number between 0 (impossibility) and 1 (certainty). In our exercise, we deal with probabilities of two events, A and B:

• The probability of event A, denoted as \(P(A)\), was given as \(\frac{1}{3}\).
• The probability of event B, denoted as \(P(B)\), was given as \(\frac{1}{4}\).

Knowing these values leads us to explore how one event might affect the likelihood of another. This relationship between two events calls for the use of conditional probability.

Bayes' Theorem

Bayes' Theorem is a foundational concept in probability that describes how to update our belief about a hypothesis based on new evidence. In simple terms, it calculates the reverse conditional probability. If you know \(P(A \mid B)\) and need \(P(B \mid A)\), Bayes’ Theorem becomes extremely useful.
Though not explicitly used in our exercise, this theorem relates to:

• \(P(B \mid A)\): The probability of B occurring given A is true.
• \(P(A \mid B)\): The probability of A occurring given B is true, provided as \(\frac{1}{6}\).

Bayes' Theorem formulates the relationship between these probabilities as:\[P(B \mid A) = \frac{P(A \mid B) \cdot P(B)}{P(A)}\]
In our solution, while the direct use of Bayes' Theorem wasn't shown, the principle behind obtaining \(P(B \mid A)\) from known values embodies the essence of this theorem.

Intersection of Events

The **intersection of events** plays a critical role when we deal with probabilities involving multiple events. The intersection of events A and B, denoted as \(P(A \cap B)\), symbolizes the probability that both events A and B happen at the same time.
In our exercise, calculating \(P(A \cap B)\) was key to finding \(P(B \mid A)\). You might wonder how this is different from regular probability:
- The intersection considers both events happening together.- We used \(P(A \mid B)\) and \(P(B)\) to calculate \(P(A \cap B)\) in our solution by rearranging the definition of conditional probability: \[P(A \cap B) = P(A \mid B) \cdot P(B)\]
This calculated intersection, \(\frac{1}{24}\), was then utilized to find \(P(B \mid A)\) using the conditional probability formula and tailored our solution towards solving the given exercise effectively.