The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respec

Q: Wait, why do we add the triple intersection to the 'exactly two' probability to get 'at least two'? Shouldn't it be separate?

That is a fantastic question! Think of 'at least two' as a bucket. If you pass in two subjects, you are in the bucket. If you pass in three subjects, you are also in the bucket because three is 'at least two'. So, the 'at least two' region is the union of the 'exactly two' region and the 'exactly three' region. They are mutually exclusive, so we simply add them.

Q: Why is the formula for 'exactly two' written as $(mp + pc + cm) - 3mpc$? Where does the 3 come from?

Great catch! When we sum the pairwise intersections ($mp + pc + cm$), we are counting the central region (where all three overlap) three times—once inside $mp$, once inside $pc$, and once inside $cm$. Since we want 'exactly two', we need to remove that central region entirely. Because it was counted three times, we must subtract it three times to clear it out completely.

Q: Is there a simpler way to solve this without the full Inclusion-Exclusion Principle?

While you could try to define variables for every single region of the Venn diagram (like $x$ for exactly one, $y$ for exactly two, etc.), the Inclusion-Exclusion Principle is the most elegant and direct path. It connects the individual probabilities to the union in one clean sweep. Trust the formula—it is designed to handle this exact complexity!

Question

The probabilities that a student passes in Mathematics, Physics and Chemistry are $m, p$ and $c$, respectively. Of these subjects, the student has a $75\%$ chance of passing in at least one, a $50\%$ chance of passing in at least two, and a $40\%$ chance of passing in exactly two. Which of the following relations are true?

Options:

(A) $p + m + c = 19/20$

(B) $p + m + c = 27/20$

(C) $pmc = 1/10$

(D) $pmc = 1/4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a probability problem; we are mapping the landscape of human success across three distinct domains: Mathematics ($M$), Physics ($P$), and Chemistry ($C$).

Imagine you are standing before a massive Venn diagram. The beauty of this problem lies in the overlap—the regions where these circles intersect. This is where the magic happens.

### The Anatomy of 'At Least'

When the problem states there is a $75\%$ chance of passing in 'at least one' subject, it defines the total area covered by the union of all three circles:
$$P(M \cup P \cup C) = 0.75$$

The problem then introduces a more restrictive condition: a $50\%$ chance of passing in 'at least two'. This means passing in exactly two subjects OR passing in all three.

Mathematically, we express this as:
$$P(	ext{At least 2}) = P(	ext{Exactly 2}) + P(	ext{Exactly 3})$$

Given $P(	ext{At least 2}) = 0.50$ and $P(	ext{Exactly 2}) = 0.40$, we find the probability of passing in all three subjects:
$$P(M \cap P \cap C) = 0.50 - 0.40 = 0.10$$

Since the events are independent, we know:
$$P(M \cap P \cap C) = m \cdot p \cdot c = \frac{1}{10}$$

### The Overcounting Trap

Now, we calculate the sum of the pairwise intersections: $mp + pc + cm$. The probability of 'exactly two' is not simply the sum of the pairwise intersections.

If you look at the Venn diagram, each pairwise intersection (e.g., $mp$) includes the central region where all three circles overlap. If we add $mp + pc + cm$, we count that central region three times.

To isolate 'exactly two', we must subtract the central region three times:
$$P(	ext{Exactly 2}) = (mp + pc + cm) - 3(mpc)$$

Substituting our known values $0.40 = (mp + pc + cm) - 3(0.10)$, we solve for the sum of pairwise intersections:
$$mp + pc + cm = 0.40 + 0.30 = 0.70$$

### The Grand Unification

Finally, we invoke the **Inclusion-Exclusion Principle**, the master key for three-set problems:
$$P(M \cup P \cup C) = (m + p + c) - (mp + pc + cm) + mpc$$

We now substitute our known values into the equation:
$$0.75 = (m + p + c) - 0.70 + 0.10$$

Simplifying the right side:
$$0.75 = (m + p + c) - 0.60$$
$$m + p + c = 1.35$$

Converting this to a fraction:
$$m + p + c = \frac{135}{100} = \frac{27}{20}$$

We have successfully navigated the logic. We found **$p + m + c = \frac{27}{20}$** and **$pmc = \frac{1}{10}$**. Probability is not just about numbers; it is about understanding the structure of overlapping realities.

Select Answer:

Visualized Solution

Visualizing the Problem with Venn Diagrams

Translating "At Least One"

Analyzing "At Least Two" vs "Exactly Two"

Calculating $P (M \cap P \cap C)$

Formula for Exactly Two Events

Calculating $\sum m p$

The Three-Set Union Formula

Solving for $m + p + c$

Converting to Fractions and Selecting Options

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Anatomy of 'At Least'

The Overcounting Trap

The Grand Unification

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