Two students A and B appeared in an examination. The probability that A will qualify the examination is 0.05 and that B will qualify the examination is 0.1. The probability that both will qualify the examination is 0.02. Find the probability that both A and B will not qualify the examination.
0.87
Understanding the Probability Problem: Students and Exam Qualification
This problem involves calculating probabilities related to two events: student A qualifying an examination and student B qualifying the same examination. We are given the individual probabilities and the probability of both events happening together.
Defining the Events and Given Probabilities
Let's define the events clearly:
- Let A be the event that student A qualifies the examination.
- Let B be the event that student B qualifies the examination.
We are given the following probabilities:
- Probability that A qualifies, \(P(A) = 0.05\)
- Probability that B qualifies, \(P(B) = 0.1\)
- Probability that both A and B qualify, \(P(A \cap B) = 0.02\)
Finding the Probability of Neither Qualifying
We need to find the probability that neither student A nor student B will qualify the examination. This is the probability that event A does not happen AND event B does not happen. In probability notation, this is \(P(A' \cap B')\), where \(A'\) is the event that A does not qualify, and \(B'\) is the event that B does not qualify.
We can use De Morgan's Law from set theory and probability, which states that \((A \cup B)' = A' \cap B'\). Therefore, the probability that neither A nor B qualifies is equal to the probability that the event "A or B (or both) qualify" does not happen. Mathematically, \(P(A' \cap B') = P((A \cup B)')\).
The probability of an event not happening is \(1\) minus the probability of the event happening. So, \(P((A \cup B)') = 1 - P(A \cup B)\).
First, let's find the probability that A or B (or both) qualify, \(P(A \cup B)\). We use the formula for the probability of the union of two events:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
Calculating \(P(A \cup B)\)
Substitute the given values into the formula:
\(P(A \cup B) = 0.05 + 0.1 - 0.02\)
\(P(A \cup B) = 0.15 - 0.02\)
\(P(A \cup B) = 0.13\)
So, the probability that at least one of the students qualifies is \(0.13\).
Calculating \(P(A' \cap B')\)
Now, we can find the probability that neither qualifies using the relationship \(P(A' \cap B') = 1 - P(A \cup B)\):
\(P(A' \cap B') = 1 - 0.13\)
\(P(A' \cap B') = 0.87\)
Thus, the probability that both A and B will not qualify the examination is \(0.87\).
Let's summarise the steps:
- Identify the given probabilities: \(P(A)\), \(P(B)\), and \(P(A \cap B)\).
- Determine the required probability: \(P(A' \cap B')\).
- Use the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) to calculate \(P(A \cup B)\).
- Use the relationship \(P(A' \cap B') = 1 - P(A \cup B)\) (derived from De Morgan's Law and complement rule) to find the final probability.
Applying these steps:
- \(P(A) = 0.05\)
- \(P(B) = 0.1\)
- \(P(A \cap B) = 0.02\)
- \(P(A \cup B) = 0.05 + 0.1 - 0.02 = 0.13\)
- \(P(A' \cap B') = 1 - 0.13 = 0.87\)
The probability that both A and B will not qualify the examination is 0.87.
Revision Table: Key Probability Concepts
| Concept | Notation | Formula/Explanation |
|---|---|---|
| Probability of Event A | \(P(A)\) | Measure of likelihood of A occurring. |
| Probability of A and B | \(P(A \cap B)\) | Probability that both A and B occur. |
| Probability of A or B (or both) | \(P(A \cup B)\) | Probability that A occurs, or B occurs, or both occur. \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). |
| Probability of Not A | \(P(A')\) or \(P(A^c)\) | Probability that A does not occur. \(P(A') = 1 - P(A)\). |
| Probability of Neither A Nor B | \(P(A' \cap B')\) | Probability that A does not occur AND B does not occur. By De Morgan's Law, \(P(A' \cap B') = P((A \cup B)')\). |
| De Morgan's Laws | \((A \cup B)' = A' \cap B'\) and \((A \cap B)' = A' \cup B'\). Useful for relating unions and intersections of complements. |
Additional Information: Set Theory and Probability Connection
Probability theory is built upon the foundations of set theory. Events are treated as sets, and the sample space is the universal set. The operations from set theory have direct counterparts in probability:
- Union (\(\cup\)): Corresponds to "OR". \(A \cup B\) is the event that A occurs OR B occurs (or both).
- Intersection (\(\cap\)): Corresponds to "AND". \(A \cap B\) is the event that both A and B occur.
- Complement (\(A'\) or \(A^c\)): Corresponds to "NOT". \(A'\) is the event that A does not occur.
Understanding these connections helps in applying formulas like the union formula and De Morgan's laws to solve probability problems involving multiple events, such as calculating the probability of neither student qualifying the exam.
For independent events, \(P(A \cap B) = P(A) * P(B)\). However, this problem provides \(P(A \cap B)\) directly, and \(0.02 \neq 0.05 * 0.1 = 0.005\), indicating that the events of A and B qualifying are NOT independent. Therefore, using the general formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) is essential.
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If A is an open set and B is a closed set, then B - A is
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