Problem 3 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(P(A)=2 / 3, P(B)=1 / 2\) and \(P(A \cup B)=5 / 6\) then events \(A\) and \(B\) are (a) mutuallye xclusive (b) independent as well as mutually exclusive (c) independent (d) dependent only on \(A\)

Step-by-Step Solution

Answer

The events \(A\) and \(B\) are dependent.

Step 1 - Define Mutually Exclusive and Independent Events

Two events are mutually exclusive if they cannot occur at the same time, i.e., their intersection is zero: \( P(A \cap B) = 0 \). Two events are independent if the occurrence of one does not affect the probability of the other, mathematically expressed as \( P(A \cap B) = P(A) \cdot P(B) \).

Step 2 - Formula for Union of Two Events

The probability of the union of two events \(A\) and \(B\) is given by \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). We will use this to find \(P(A \cap B)\).

Step 3 - Calculate \(P(A \cap B)\)

Substitute the given probabilities into the formula for the union: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \].\[ \frac{5}{6} = \frac{2}{3} + \frac{1}{2} - P(A \cap B) \].Simplifying: \[ \frac{5}{6} = \frac{4}{6} + \frac{3}{6} - P(A \cap B) \] leads to \[ P(A \cap B) = \frac{1}{6} \].

Step 4 - Evaluate Conditions for Exclusivity and Independence

Since \( P(A \cap B) eq 0 \), events \( A \) and \( B \) are not mutually exclusive. For independence, check if \( P(A \cap B) = P(A) \cdot P(B) \):\[ \frac{1}{6} = \frac{2}{3} \cdot \frac{1}{2} \].Calculate:\[ \frac{1}{6} = \frac{1}{3} \].Since this is not true, the events are not independent.

Key Concepts

Mutually Exclusive Events

Mutually exclusive events are those that cannot happen at the same time. Think of it as having to choose between wearing a red shirt or a blue shirt. You can't wear both at once; picking one excludes the other.

In probability terms, if two events are mutually exclusive, the probability that both events occur is zero. Mathematically, we write this as:

• \( P(A \cap B) = 0 \)

So anytime you're told events might be mutually exclusive, you should check if this statement holds true.
Imagine events A and B could happen at a university fair where A is attending a seminar and B is visiting a robotics booth. If these seminars and booths happen simultaneously and in different parts of the campus, attending both isn't feasible at the same time.

This exercise asked if events A and B were mutually exclusive. Initially, using \( P(A \cap B) = \frac{1}{6} \), we see they are not. Since that probability is more than zero, the events can occur together, meaning they are not mutually exclusive.

Independent Events

Independent events describe situations where the occurrence or outcome of one event doesn't affect the other. Picture flipping two separate coins. Whether the first lands heads or tails has no bearing on the result of the second coin flip.

In probability terms, if two events are independent, the probability of both events occurring is the product of their individual probabilities. Mathematically, it is expressed by:

• \( P(A \cap B) = P(A) \cdot P(B) \)

To determine if events are independent, you can calculate both sides of the above equation and compare.

In our exercise, substituting the values, we checked if \( \frac{1}{6} = \frac{2}{3} \cdot \frac{1}{2} \) holds. Simplifying both sides, it turns out \( \frac{1}{6} = \frac{1}{3} \), which is not true. Thus, events A and B in the scenario are not independent either. Understanding this concept helps when assessing linked probabilities, ensuring one event doesn't inadvertently sway another's likelihood.

Probability of Union

The probability of union brings together the chances of either event occurring, focusing on inclusivity rather than exclusivity. Suppose you want either cake or ice cream for dessert; this gives the opportunity to enjoy both together or just one. Using probabilities, it is given by the formula:

• \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

This calculation involves adding the probabilities of each individual event and then subtracting the overlap, where both occur simultaneously to avoid double-counting.

Using the given data, \( P(A) = \frac{2}{3}, \) \( P(B) = \frac{1}{2}, \) and \( P(A \cup B) = \frac{5}{6} \), we apply the formula: - \( \frac{5}{6} = \frac{2}{3} + \frac{1}{2} - P(A \cap B) \)
Simplifying, you'll find \( P(A \cap B) = \frac{1}{6} \).
The exercise clearly illustrates how understanding this formula saves you from misinterpreting overlaps, keeping separately counting probabilities clear and accurate. Knowing when to apply the probability of union helps determine how often you can expect either or both events to occur in probability puzzles.