Understanding the Problem of Student Test Probability

The problem asks us to find a relationship between the probabilities of a student passing three different tests, given specific conditions for success and the overall probability of success. We are given the probabilities of passing test I, test II, and test III as m, n, and 1/2, respectively.

The student is considered successful under the following conditions:

• Passes in test I and test II.
• Passes in test I and test III.
• Passes in all three tests (I, II, and III).

We are also given that the total probability of the student being successful is 1/2. We need to use this information to determine the correct relationship between m and n from the given options.

Defining Events and Probabilities

Let's define the events of passing each test:

• Let \(A\) be the event that the student passes test I.
• Let \(B\) be the event that the student passes test II.
• Let \(C\) be the event that the student passes test III.

From the problem statement, we have the probabilities of these events:

• \(P(A) = m\)
• \(P(B) = n\)
• \(P(C) = \frac{1}{2}\)

We assume that the outcomes of the three tests are independent events. This means the result of one test does not affect the result of another.

Identifying the Success Conditions

The student is successful if any of the following composite events occur:

• The student passes test I AND test II (\(A \cap B\)).
• The student passes test I AND test III (\(A \cap C\)).
• The student passes test I AND test II AND test III (\(A \cap B \cap C\)).

Let \(S\) be the event that the student is successful. The event \(S\) can be expressed as the union of the above conditions:

\(S = (A \cap B) \cup (A \cap C) \cup (A \cap B \cap C)\)

Notice that if the student passes all three tests (\(A \cap B \cap C\)), they have also passed test I and test II (\(A \cap B\)), and they have also passed test I and test III (\(A \cap C\)). Therefore, the condition \(A \cap B \cap C\) is already included within the first two conditions.

So, the event of success simplifies to:

\(S = (A \cap B) \cup (A \cap C)\)

Calculating the Probability of Success

We need to find the probability of the union of two events, \((A \cap B)\) and \((A \cap C)\). Using the principle of inclusion-exclusion for two events \(X\) and \(Y\), \(P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)\).

Here, \(X = A \cap B\) and \(Y = A \cap C\). The intersection of \(X\) and \(Y\) is \((A \cap B) \cap (A \cap C) = A \cap B \cap C\).

So, the probability of success is:

\(P(S) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)\)

Since the tests are independent, the probability of the intersection of events is the product of their individual probabilities:

• \(P(A \cap B) = P(A) \times P(B) = m \times n = mn\)
• \(P(A \cap C) = P(A) \times P(C) = m \times \frac{1}{2} = \frac{m}{2}\)
• \(P(A \cap B \cap C) = P(A) \times P(B) \times P(C) = m \times n \times \frac{1}{2} = \frac{mn}{2}\)

Substitute these probabilities into the equation for \(P(S)\):

\[P(S) = mn + \frac{m}{2} - \frac{mn}{2}\]

Combine the terms involving \(mn\):

\[P(S) = \frac{m}{2} + \left(mn - \frac{mn}{2}\right)\]

\[P(S) = \frac{m}{2} + \frac{mn}{2}\]

Finding the Relationship between m and n

We are given that the probability of the student being successful is \(P(S) = \frac{1}{2}\). Set the derived expression for \(P(S)\) equal to this value:

\[\frac{1}{2} = \frac{m}{2} + \frac{mn}{2}\]

To eliminate the denominators, multiply the entire equation by 2:

\[2 \times \left(\frac{1}{2}\right) = 2 \times \left(\frac{m}{2} + \frac{mn}{2}\right)\]

\[1 = m + mn\]

Now, factor out the common term \(m\) on the right side of the equation:

\[1 = m(1 + n)\]

This gives us the required relationship between \(m\) and \(n\).

Comparing with the Options

Let's compare the derived relationship \(m(1 + n) = 1\) with the given options:

The derived relationship \(m(1 + n) = 1\) matches Option 1.

Revision Table: Key Probability Concepts

Additional Information: Probability in Test Scenarios

Probability problems involving test outcomes are common and often rely on understanding how to combine probabilities for independent or dependent events. In this case, assuming independence simplified the calculations significantly. If the tests were dependent (e.g., passing test I makes it easier to pass test II), we would need conditional probabilities \(P(B|A)\) or \(P(C|A)\) to calculate the joint probabilities \(P(A \cap B)\), \(P(A \cap C)\), and \(P(A \cap B \cap C)\). The inclusion-exclusion principle is crucial when dealing with the probability of the union of multiple events, especially when they can overlap (i.e., are not mutually exclusive). Identifying the simplified success event \(S = (A \cap B) \cup (A \cap C)\) was key to solving this problem efficiently.