Problem 132
Question
If \(A\) and \(B\) are two events, then the probability that at most one of \(A, B\) occurs is (A) \(1-P(A \cap B)\) (B) \(P\left(A^{\prime}\right)+P\left(B^{\prime}\right)-P\left(A^{\prime} \cap B^{\prime}\right)\) (C) \(P\left(A^{\prime}\right)+P\left(B^{\prime}\right)+P(A \cup B)\) (D) none of these
Step-by-Step Solution
Verified Answer
The probability that at most one of \(A\) or \(B\) occurs is (A) \(1 - P(A \cap B)\).
1Step 1: Identify the Given Information
We are given two events, \(A\) and \(B\). Our goal is to find the probability that at most one of these events occurs.
2Step 2: Express At Most One Occurrence
The event 'at most one of \(A\) or \(B\) occurs' includes the following scenarios: neither \(A\) nor \(B\) occurs, or exactly one of the events \(A\) or \(B\) occurs. Mathematically, this can be expressed as \(A^{\prime} \cap B^{\prime} \) or \((A \cap B^{\prime}) \cup (A^{\prime} \cap B)\).
3Step 3: Calculate Probability Components
The probability of \(A^{\prime} \cap B^{\prime}\) is the chance that neither \(A\) nor \(B\) occurs. The probabilities for exactly one event occurring are \(P(A) - P(A \cap B)\) for event \(A\) only, and \(P(B) - P(A \cap B)\) for event \(B\) only.
4Step 4: Use Probability Addition Rule
According to the addition rule for probabilities, for mutually exclusive events, we sum the probabilities. Thus, the probability of exactly one or no events occurring is: \[ P(A^{\prime} \cap B^{\prime}) + (P(A) - P(A \cap B)) + (P(B) - P(A \cap B)) \].
5Step 5: Simplify the Expression
Using the formula for total probability, simplify: \(P(A^{\prime} \cap B^{\prime}) = 1 - P(A \cup B)\), then \((P(A) - P(A \cap B)) + (P(B) - P(A \cap B)) = P(A) + P(B) - 2P(A \cap B)\). Then the total simplifies further to: \[1 - P(A \cap B)\].
6Step 6: Select the Correct Answer
From the choices, the expression \(1 - P(A \cap B)\) corresponds to option (A).
Key Concepts
Events and OutcomesAddition Rule for ProbabilitiesMutually Exclusive Events
Events and Outcomes
In probability theory, an **event** is a specific outcome or a set of outcomes of a random process. Each possible event in the sample space has an associated probability. These probabilities determine how likely each event is to occur.
- An **outcome** refers to a result of a single trial of a random process. For example, getting a 'heads' when you flip a coin is an outcome.
- Multiple outcomes together make an event. For instance, rolling an even number on a die (2, 4, or 6) is an event made of three outcomes.
Addition Rule for Probabilities
The **addition rule for probabilities** is a key concept in probability that helps us calculate the chances of either one event or another happening. This rule is particularly helpful when events have overlapping outcomes.The basic form of the rule when events are not mutually exclusive is given by:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]
- **\( P(A \cup B) \)**: Probability that either event **A** or event **B** or both occur.
- **\( P(A) \) & \( P(B) \)**: Individual probabilities of events **A** and **B** occurring.
- **\( P(A \cap B) \)**: Probability that both events occur simultaneously. We subtract this to adjust for the overlap, as it was counted twice.
Mutually Exclusive Events
**Mutually exclusive events** are outcomes that cannot occur simultaneously in a single trial. This concept is crucial in correctly calculating probabilities for events that do not have overlapping outcomes.Here are some characteristics:
- If events **A** and **B** are mutually exclusive, the probability of both occurring at the same time is 0: \( P(A \cap B) = 0 \).
- The addition rule simplifies for mutually exclusive events to \( P(A \cup B) = P(A) + P(B) \) since there is no overlap.
Other exercises in this chapter
Problem 127
If \(A\) and \(B\) are two events such that \(P(A)=\frac{1}{2}\) and \(P(B)=\frac{2}{3}\), then (A) \(P(A \cup B) \geq \frac{2}{3}\) (B) \(P\left(A \cap B^{\pri
View solution Problem 131
If \(P(A)=\frac{2}{5}\) and \(P(B)=\frac{4}{5}\), then (A) \(P(A \cup B) \geq \frac{4}{5}\) (B) \(\frac{1}{5} \leq P(A \cap B) \leq \frac{2}{5}\) (C) \(\frac{1}
View solution Problem 142
Assertion: \(A\) set \(X\) contains \(n\) elements. Two subsets \(A\) and \(B\) of \(X\) are chosen at random. The probability that \(A\) and \(B\) have same nu
View solution Problem 143
Assertion: A bag contains \(n+1\) coins. It is known that one of these coins has a head on both sides while the other coins are fair. One coin is selected at ra
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