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Official PYQ · 2015Set 1 · QGA-10MCQ1 mark

GATE CS 2015 Set 1 QGA-10 — General Aptitude

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The probabilities that a student passes in mathematics, physics and chemistry are m,pm,p and cc respectively. Of these subjects, the student has 75%75\% chance of passing in at least one, a 50%50\% chance of passing in at least two and a 40%40\% chance of passing in exactly two. Following relations are drawn in m,p,cm, p, c:

  1. p+m+c=27/20p + m + c = 27/20
  2. p+m+c=13/20p + m + c = 13/20
  3. (p)×(m)×(c)=1/10(p) \times (m) \times (c) = 1/10
Choose one option
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Answer: D

Let x1,x2,x_1, x_2, and x3x_3 denote the probabilities of the student passing in exactly one, exactly two, and exactly three subjects, respectively.

From the given information:

  • Passing in at least one subject: x1+x2+x3=0.75x_1 + x_2 + x_3 = 0.75
  • Passing in at least two subjects: x2+x3=0.50x_2 + x_3 = 0.50
  • Passing in exactly two subjects: x2=0.40x_2 = 0.40

Solving these equations, we get:

  • x3=0.50−0.40=0.10x_3 = 0.50 - 0.40 = 0.10
  • x1=0.75−0.50=0.25x_1 = 0.75 - 0.50 = 0.25

Using the principle of inclusion-exclusion, the sum of individual probabilities is given by:

m+p+c=x1+2x2+3x3m + p + c = x_1 + 2x_2 + 3x_3 m+p+c=0.25+2(0.40)+3(0.10)=1.35=2720m + p + c = 0.25 + 2(0.40) + 3(0.10) = 1.35 = \frac{27}{20}

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

Assuming the events of passing the subjects are independent, the probability of passing in all three subjects is:

m×p×c=x3=0.10=110m \times p \times c = x_3 = 0.10 = \frac{1}{10}

Thus, Relation III is true.

Therefore, Relations I and III are true, which corresponds to Option D.

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