GATE CS 2015 Set 1 Q.10 (GA Q.10): Quantitative Aptitude

The probabilities that a student passes in Mathematics, Physics and Chemistry are $m, p,$ and $c$ respectively. Of these subjects, the student has $75\%$ 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. Following relations are drawn in $m, p, c$ :

\begin{aligned} \text{(I)} & \quad p + m + c = 27/20 \\ \text{(II)} & \quad p + m + c = 13/20 \\ \text{(III)} & \quad (p) \times (m) \times (c) = 1/10 \end{aligned}

Which of the following options is correct?

Only relation I is true.

Only relation II is true.

Relations II and III are true.

Relations I and III are true.

Official answer and solution

Official answer: D
From: Official GATE 2015 answer key

Solution

Apply the Principle of Inclusion-Exclusion for three events to relate marginal probabilities to intersection probabilities.

Let $M, P, C$ denote passing Mathematics, Physics, and Chemistry, with individual probabilities $P(M) = m, P(P) = p, P(C) = c$ .

Given:

$P(\text{at least one}) = P(M \cup P \cup C) = 0.75$ .
$P(\text{at least two}) = 0.50$ .
$P(\text{exactly two}) = 0.40$ .

Find $P(\text{all three}) = P(M \cap P \cap C)$ :
Since passing at least two equals passing exactly two plus passing all three:

P(\text{at least two}) = P(\text{exactly two}) + P(M \cap P \cap C)

0.50 = 0.40 + P(M \cap P \cap C) \implies P(M \cap P \cap C) = 0.10 = \frac{1}{10}.

Under the standard assumption of independent examinations across subjects, $P(M \cap P \cap C) = p \cdot m \cdot c = \frac{1}{10}$ . Hence, Relation (III) is true.

Find $p + m + c$ using Inclusion-Exclusion:

P(M \cup P \cup C) = (m + p + c) - \sum P(A \cap B) + P(M \cap P \cap C)

Note that every pairwise intersection $A \cap B$ includes the triple intersection $M \cap P \cap C$ . Thus:

\sum P(A \cap B) = P(\text{exactly two}) + 3 P(M \cap P \cap C) = 0.40 + 3(0.10) = 0.70.

Substituting into the inclusion-exclusion formula:

0.75 = (m + p + c) - 0.70 + 0.10

0.75 = (m + p + c) - 0.60

m + p + c = 0.75 + 0.60 = 1.35 = \frac{27}{20}.

Hence, Relation (I) is true and Relation (II) is false.

Therefore, Relations I and III are true (option D).

Final answer

Common trap

Forgetting that the sum of pairwise intersections $\sum P(A \cap B)$ counts the triple intersection three times, which would lead to an incorrect value for $m + p + c$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 1
Question: 10 of 65, printed as GA Q.10
Section: General Aptitude

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