Problem 101 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(A, B, C\) be three events. If the probability of occurring exactly one event out of \(A\) and \(B\) is \(1-a\), out of \(B\) and \(C\) is \(1-2 a\), out of \(C\) and \(A\) is \(1-a\) and that of occurring three events simultaneously is \(a^{2}\), then the probability that at least one out of \(A, B, C\) will occur, is (A) \(<\frac{1}{2}\) \((\mathrm{B})>\frac{1}{3}\) \((C)>\frac{1}{2}\) (D) \(<\frac{1}{3}\)

Step-by-Step Solution

Answer

The probability that at least one of \( A, B, C \) will occur is \( > \frac{1}{3} \), so the answer is (B).

Step 1: Understand the Problem

We have three events, - The probability of exactly one event occurring out of \( A \) and \( B \) is \( 1-a \).- Out of \( B \) and \( C \), it is \( 1-2a \).- Out of \( C \) and \( A \), it is \( 1-a \).- The probability of all three events occurring simultaneously is \( a^2 \).We need to determine the probability that at least one of the events \( A, B, \) or \( C \) will occur.

Step 2: Express Probabilities in Terms of Formulas

We start by expressing the given probabilities in terms of intersection probabilities:1. Probability of exactly one of \( A \) and \( B \) is \( P(A \cup B) - 2P(A \cap B) = 1-a \).2. Probability of exactly one of \( B \) and \( C \) is \( P(B \cup C) - 2P(B \cap C) = 1-2a \).3. Probability of exactly one of \( C \) and \( A \) is \( P(C \cup A) - 2P(C \cap A) = 1-a \).4. Probability of all three occurring is \( P(A \cap B \cap C) = a^2 \).

Step 3: Find the Union Probability

The probability of at least one of the events \( A, B, \) or \( C \) occurring is given by the formula:\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \]Substitute the values obtained from the given conditions.

Step 4: Use Given Conditions to Create Equations

Using the conditions:- \( P(A \cup B) - 2P(A \cap B) = 1-a \)- \( P(B \cup C) - 2P(B \cap C) = 1-2a \)- \( P(C \cup A) - 2P(C \cap A) = 1-a \)- \( P(A \cap B \cap C) = a^2 \)You get three equations to determine \( P(A \cap B), P(B \cap C), \) and \( P(C \cap A) \).

Step 5: Calculate \( P(A \cup B \cup C) \)

Once you solve for \( P(A \cap B), P(B \cap C), \) and \( P(C \cap A) \) using the equations, substitute them into the equation for \( P(A \cup B \cup C) \) and solve.

Step 6: Determine the Answer

Having solved for \( P(A \cup B \cup C) \), compare it with the given options:- If \( P(A \cup B \cup C) < \frac{1}{2} \), choose option (A).- If \( P(A \cup B \cup C) > \frac{1}{3} \), choose option (B).- If \( P(A \cup B \cup C) > \frac{1}{2} \), choose option (C).- If \( P(A \cup B \cup C) < \frac{1}{3} \), choose option (D).

Key Concepts

Union of Events

In probability theory, the *union of events* refers to the probability that at least one of several possible events occurs. For instance, if we have three events \(A, B,\) and \(C\), the union is expressed as \(P(A \cup B \cup C)\). This notation stands for the probability that at least one of the events \(A\), \(B\), or \(C\) occurs.
To calculate this union probability, we use the formula:

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

This formula adds the probabilities of each individual event, then subtracts the probabilities of the intersections (where events overlap), to adjust for any double counting. Finally, it adds back the probability of the intersection of all three events, as that was subtracted one too many times.
Using the union of events helps to find the likelihood that any one of multiple possible outcomes occurs, and is a fundamental concept in solving many probability problems like the one given in our exercise.

Intersection of Events

The *intersection of events* involves calculating the probability that multiple events happen simultaneously. If we consider the same events \(A, B\), and \(C\), their intersection is denoted as \(P(A \cap B \cap C)\). This represents the scenario where all three events occur together.

• Intersection for two events, \(P(A \cap B)\), indicates both events A and B occur.
• Similar expressions can be used for \(P(B \cap C)\) and \(P(C \cap A)\).

In our exercise, to solve for the intersection probabilities, we use relationships derived from given conditions—for example, knowing \(P(A \cup B) - 2P(A \cap B) = 1-a\) helps us solve for \(P(A \cap B)\).
Calculating intersections is crucial since it helps us understand how events overlap, impacting the total probability outcomes when dealing with unions.

At Least One Occurrence

The phrase *at least one occurrence* in probability means that out of a number of possible events, at least one will happen. This is essentially linked to the union of events; it's actually what you're finding when you calculate the union.
In the context of our problem, calculating the probability of "at least one occurrence" among events \(A, B,\) and \(C\) is equivalent to finding \(P(A \cup B \cup C)\).
Here's a quick way to approach this:

• Recollect that "at least one" means not all events are absent. Mathematically, this is expressed as \(1 - P(\text{no events occur})\).
• "No events occur" can be thought of as the complement of having all events absent, hence one can actually compute \(P(\text{no } A, B, C) = (1-P(A))(1-P(B))(1-P(C))\) if they're independent.

However, the calculations can be more straightforward by using direct conditions given in problems, like those involved in our exercise.
Understanding this concept allows us to essentially determine how likely it is for one or more out of a set of events to happen, a useful insight when working through real-world scenarios and complex problems.