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If \(A\) and \(B\) are two events such that \(P(A \cup B) \geq \frac{3}{4}\) and \(\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}\), then (A) \(P(A)+P(B) \leq \frac{11}{8}\) (B) \(P(A) \cdot P(B) \leq \frac{3}{8}\) (C) \(P(A)+P(B) \geq \frac{7}{8}\) (D) none of these

Short Answer

Expert verified
The correct answer is (A) \(P(A)+P(B) \leq \frac{11}{8}\).

Step by step solution

01

Identify Given Conditions

We are given that \(P(A \cup B) \geq \frac{3}{4}\) and \(\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}\). These inequalities will help us determine the relationship between \(P(A)\) and \(P(B)\).
02

Apply Formula for Union of Events

The formula for the probability of the union of two events is \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Rearranging it gives us \(P(A) + P(B) = P(A \cup B) + P(A \cap B)\).
03

Substitute Known Inequalities

Substitute the known inequalities into the rearranged formula: \(P(A) + P(B) = P(A \cup B) + P(A \cap B)\). Using the least values, \(P(A) + P(B) \geq \frac{3}{4} + \frac{1}{8} = \frac{7}{8}\).
04

Evaluate the Other Inequality

Using the greatest values gives \(P(A) + P(B) \leq P(A \cup B) + P(A \cap B) \leq 1 + \frac{3}{8} = \frac{11}{8}\).
05

Assess Options Against Results

From our analysis, both (A) and (C) appear to be true based on different substitutions. However, option (A) is directly confirmed by \(P(A) + P(B) \leq \frac{11}{8}\).
06

Conclusion

Comparing our results to the options shows that option (A) \(P(A)+P(B) \leq \frac{11}{8}\) is consistent with the calculated inequalities.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Union of Events
In probability theory, the concept of the "Union of Events" refers to the event that at least one of several events occurs. When we talk about the union of two events, say \(A\) and \(B\), we are considering any scenario where either event \(A\) happens, event \(B\) happens, or both events happen simultaneously.

Mathematically, the probability of the union is denoted as \(P(A \cup B)\). The formula to calculate this probability is given by:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula accounts for the basic addition of probabilities but subtracts the intersection part \(P(A \cap B)\) because it gets counted twice when we simply add \(P(A)\) and \(P(B)\).

In our exercise, we know that \(P(A \cup B) \geq \frac{3}{4}\), which is a significant piece of information. This tells us that there's a high probability that either event \(A\), event \(B\), or both happen.
Intersection of Events
"Intersection of Events" focuses on the scenario where two events happen simultaneously. In probability terms, if we consider events \(A\) and \(B\), the intersection, denoted as \(P(A \cap B)\), measures the probability that both \(A\) and \(B\) occur together.

The range for \(P(A \cap B)\) in our exercise is given as \(\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}\). This range is essential because it provides limits within which the probability of both events happening lies.

Understanding this concept ensures that when analyzing the overall probabilities involving \(A\) and \(B\), we can accurately adjust for when they're mutually dependent or overlap. In simpler terms, when both events occur, this value reduces the over-counting in the union formula.
Probability Inequality
Probability inequalities provide us with bounds and relationships between different probabilities. They help in estimating or comparing probabilities when exact figures are not possible to determine.

In the given exercise, we see examples of probability inequalities. For instance:
  • \(P(A \cup B) \geq \frac{3}{4}\)
  • \(\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}\)
By rearranging and substituting these inequalities in the union and intersection formulas, we derive larger or smaller bounds for different probabilities, such as \(P(A) + P(B)\).

Using these inequalities helps us understand the probability landscape between events like \(A\) and \(B\) without knowing each probability in isolation. They provide insights into how probabilities are distributed across individual and joint events, guiding us to determine the plausibility of various scenarios or outcomes.

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Most popular questions from this chapter

The probability that \(A\) speaks truth is \(\frac{4}{5}\), while this probability for \(B\) is \(\frac{3}{4}\). The probability that they contradict each other when asked to speak on a fact is \([\mathbf{2 0 0 4}]\) (A) \(\frac{3}{20}\) (B) \(\frac{1}{5}\) (C) \(\frac{7}{20}\) (D) \(\frac{4}{5}\)

Plant I of \(X Y Z\) manufacturing organization employs 5 production and 3 maintenance foremen, another plant II of same organization employs 4 production and 5 maintenance foremen. From any one of these plants, a single selection of two foremen is made. The probability that one of them would be production and the other maintenance foreman is (A) \(\frac{275}{504}\) (B) \(\frac{263}{504}\) (C) \(\frac{301}{504}\) (D) \(\frac{362}{504}\)

\(A\) die is thrown. Let \(A\) be the event that the number obtained is greater than 3 . Let \(B\) be the event that the number obtained is less than 5 . then \(P(A \cup B)\) is (A) \(3 / 5\) (B) 0 (C) 1 (D) \(2 / 5\)

An unbiased die with faces marked \(1,2,3,4,5\) and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5 is (A) \(\frac{16}{81}\) (B) \(\frac{1}{81}\) (C) \(\frac{80}{81}\) (D) \(\frac{65}{81}\)

At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10 -minute time intervals. The probability that there is at the most one phone call during a 10 -minute time period is [2006] (A) \(\frac{6}{5^{c}}\) (B) \(\frac{5}{6}\) (C) \(\frac{6}{55}\) (D) \(\frac{6}{e^{5}}\)

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