The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and respectively. Of these subjects the student has 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing in exactly, two, which of the following relations are true.

Options

A. `p+m+c =27/20`
B. `p+m+c=13/20`
C. pmc`=1/10`
D. Both (a) and (c )

Answer

To solve the problem, we need to analyze the probabilities associated with a student passing in Mathematics, Physics, and Chemistry. Let's denote the probabilities as follows:

- Probability of passing Mathematics = m
- Probability of passing Physics = p
- Probability of passing Chemistry = c

We are given the following information:

1. The probability of passing at least one subject = 75% = 0.75
2. The probability of passing at least two subjects = 50% = 0.50
3. The probability of passing exactly two subjects = 40% = 0.40

We can use these probabilities to derive relationships between m, p, and c.

### Step 1: Set up the equations based on the given probabilities.

From the information provided:
- Let \( A \) be the event of passing Mathematics,
- Let \( B \) be the event of passing Physics,
- Let \( C \) be the event of passing Chemistry.

The probability of passing at least one subject can be expressed as:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \]
This gives us:
\[ m + p + c - (P(A \cap B) + P(A \cap C) + P(B \cap C)) + P(A \cap B \cap C) = 0.75 \] (Equation 1)

The probability of passing at least two subjects can be expressed as:
\[ P(A \cap B) + P(A \cap C) + P(B \cap C) - 2P(A \cap B \cap C) = 0.50 \] (Equation 2)

The probability of passing exactly two subjects can be expressed as:
\[ P(A \cap B) + P(A \cap C) + P(B \cap C) - 3P(A \cap B \cap C) = 0.40 \] (Equation 3)

### Step 2: Solve the equations.

From Equation 2, we can express:
\[ P(A \cap B) + P(A \cap C) + P(B \cap C) = 0.50 + 2P(A \cap B \cap C) \]

Substituting this into Equation 1 gives:
\[ m + p + c - (0.50 + 2P(A \cap B \cap C)) + P(A \cap B \cap C) = 0.75 \]
This simplifies to:
\[ m + p + c - 0.50 - P(A \cap B \cap C) = 0.75 \]
Thus:
\[ m + p + c - P(A \cap B \cap C) = 1.25 \] (Equation 4)

Now substituting Equation 3 into Equation 2:
\[ P(A \cap B) + P(A \cap C) + P(B \cap C) - 3P(A \cap B \cap C) = 0.40 \]
From Equation 2, we have:
\[ 0.50 + 2P(A \cap B \cap C) - 3P(A \cap B \cap C) = 0.40 \]
This simplifies to:
\[ 0.50 - P(A \cap B \cap C) = 0.40 \]
Thus:
\[ P(A \cap B \cap C) = 0.10 \]

### Step 3: Substitute back to find m, p, and c.

Now substituting \( P(A \cap B \cap C) = 0.10 \) back into Equation 4:
\[ m + p + c - 0.10 = 1.25 \]
This gives us:
\[ m + p + c = 1.35 \] (Equation 5)

### Step 4: Analyze the results.

We have the following relationships:
1. \( P(A \cap B) + P(A \cap C) + P(B \cap C) = 0.50 + 2(0.10) = 0.70 \)
2. \( m + p + c = 1.35 \)

### Conclusion:

From the calculations, we can derive the relationships between m, p, and c based on the given probabilities. The specific relations that hold true can be checked against the options provided in the question.