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Official PYQ · 2012Q63MCQ1 mark

GATE CS 2012 Q63 — General Aptitude

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An automobile plant contracted to buy shock absorbers from two suppliers XX and YY . XX supplies 60%60\% and Y supplies 40%40\% of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable. Of X’sX’s shock absorbers, 96%96\% are reliable. Of Y’sY’s shock absorbers, 72%72\% are reliable.

The probability that a randomly chosen shock absorber, which is found to be reliable, is made by YY is

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Answer: B

We can solve this problem using Bayes' Theorem.

Let us define the following events:

  • XX: The shock absorber is supplied by XX.
  • YY: The shock absorber is supplied by YY.
  • RR: The shock absorber is reliable (passes the quality test).

From the given information, we have the following probabilities:

  • P(X)=0.60P(X) = 0.60
  • P(Y)=0.40P(Y) = 0.40
  • P(R∣X)=0.96P(R|X) = 0.96
  • P(R∣Y)=0.72P(R|Y) = 0.72

We need to find the conditional probability that the shock absorber is made by YY, given that it is reliable, i.e., P(Y∣R)P(Y|R).

By Bayes' Theorem:

P(Y∣R)=P(R∣Y)⋅P(Y)P(R)P(Y|R) = \frac{P(R|Y) \cdot P(Y)}{P(R)}

First, we calculate the total probability of a shock absorber being reliable, P(R)P(R):

P(R)=P(R∣X)⋅P(X)+P(R∣Y)⋅P(Y)=(0.96×0.60)+(0.72×0.40)=0.576+0.288=0.864\begin{aligned} P(R) &= P(R|X) \cdot P(X) + P(R|Y) \cdot P(Y) \\ &= (0.96 \times 0.60) + (0.72 \times 0.40) \\ &= 0.576 + 0.288 \\ &= 0.864 \end{aligned}

Now, we compute P(Y∣R)P(Y|R):

P(Y∣R)=0.2880.864=13≈0.3333\begin{aligned} P(Y|R) &= \frac{0.288}{0.864} \\ &= \frac{1}{3} \\ &\approx 0.3333 \end{aligned}

Among the given options, 0.334 (Option B) is the closest approximation.

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