The probabilities that a student will solve Question A and Question B are 0.4 and 0.5 respectively. What is the probability that he solves at least one of the two questions?
0.7
Understanding the Probability of Solving Questions
This question asks for the probability that a student solves at least one of two questions, Question A and Question B. We are given the individual probabilities of solving each question.
Given Probabilities:
- Probability of solving Question A, denoted as $P(A) = 0.4$.
- Probability of solving Question B, denoted as $P(B) = 0.5$.
We want to find the probability of the student solving at least one of the two questions. This means the student could solve A only, solve B only, or solve both A and B. In probability terms, this is finding the probability of the union of events A and B, \(P(A \cup B)\).
Assuming the events of solving Question A and solving Question B are independent (solving one doesn't affect the probability of solving the other), we can use the formula for the union of two events or the complement rule.
Method 1: Using the Complement Rule
The event "solving at least one" is the complement of the event "solving neither". Let $A'$ be the event that the student does not solve Question A, and $B'$ be the event that the student does not solve Question B.
- Probability of not solving Question A: $P(A') = 1 - P(A) = 1 - 0.4 = 0.6$.
- Probability of not solving Question B: $P(B') = 1 - P(B) = 1 - 0.5 = 0.5$.
Since solving A and solving B are independent events, not solving A and not solving B are also independent events. The probability of solving neither A nor B is the probability of $A'$ and $B'$ occurring together, which is \(P(A' \cap B')\).
For independent events:
\( P(A' \cap B') = P(A') \times P(B') \)
Calculating the probability of solving neither question:
\( P(\text{solving neither}) = 0.6 \times 0.5 = 0.3 \)
The probability of solving at least one question is \(1\) minus the probability of solving neither question.
\( P(\text{at least one}) = 1 - P(\text{solving neither}) \)
\( P(\text{at least one}) = 1 - 0.3 = 0.7 \)
Method 2: Using the Union Formula for Independent Events
For any two events A and B, the probability of their union is given by \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). If A and B are independent, the probability of their intersection \( P(A \cap B) \) is \( P(A) \times P(B) \).
First, calculate the probability of solving both questions (A and B):
\( P(A \cap B) = P(A) \times P(B) \)
\( P(A \cap B) = 0.4 \times 0.5 = 0.2 \)
Now, use the union formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = 0.4 + 0.5 - 0.2 \)
\( P(A \cup B) = 0.9 - 0.2 \)
\( P(A \cup B) = 0.7 \)
Both methods yield the same result. The probability that the student solves at least one of the two questions is 0.7.
| Event | Probability |
|---|---|
| Solves Question A ($P(A)$) | 0.4 |
| Solves Question B ($P(B)$) | 0.5 |
| Does not solve Question A ($P(A')$) | 0.6 |
| Does not solve Question B ($P(B')$) | 0.5 |
| Solves neither (\(P(A' \cap B')\)) | \(0.6 \times 0.5 = 0.3\) |
| Solves at least one (\(P(A \cup B)\)) | $1 - 0.3 = 0.7$ |
The final answer is 0.7.
Probability Revision Table
| Concept | Description | Formula (for independent events A, B) |
|---|---|---|
| Probability of Event A | The likelihood of event A occurring. | $P(A)$ (given) |
| Complement of A | Event A does not occur ($A'$). | $P(A') = 1 - P(A)$ |
| Intersection (\(A \cap B\)) | Both events A and B occur. | \(P(A \cap B) = P(A) \times P(B)\) |
| Union (\(A \cup B\)) | At least one of events A or B occurs. | \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) or \(P(A \cup B) = 1 - P(A' \cap B')\) |
| Independent Events | The outcome of one event does not affect the outcome of the other. | \(P(A \cap B) = P(A) \times P(B)\) |
Additional Probability Information
When dealing with probability questions, it's crucial to identify whether the events are independent or dependent. In this question, it's reasonable to assume that solving one question does not impact the difficulty or the student's ability to solve the other, making them independent events.
- Independent Events: Events A and B are independent if the occurrence of A does not affect the probability of B, and vice versa. Mathematically, this means $P(B|A) = P(B)$ or $P(A|B) = P(A)$. A key property is \(P(A \cap B) = P(A) \times P(B)\).
- Dependent Events: Events are dependent if the outcome of one event influences the outcome of another. For dependent events, \(P(A \cap B) = P(A) \times P(B|A)\).
- Mutually Exclusive Events: Events are mutually exclusive if they cannot occur at the same time, meaning \(P(A \cap B) = 0\). For mutually exclusive events, \(P(A \cup B) = P(A) + P(B)\). However, independent events are generally NOT mutually exclusive unless one event has zero probability. Solving Question A and solving Question B are not mutually exclusive since the student can potentially solve both.
Understanding the concepts of union, intersection, and complements is fundamental to solving problems involving multiple events. The "at least one" type of problem is frequently solved efficiently using the complement rule, as shown in Method 1.
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