GATE CSE 2010 | Question: 26

Question

Consider a company that assembles computers. The probability of a faulty assembly of any computer is $p$. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of $q$. What is the probability of a computer being declared faulty?

• A. $pq + (1 - p)(1 - q)$
• B. $(1 - q)p$
• C. $(1 - p)q$
• D. $pq$

Comments

in this question two options ( B & C ) are same please correct it

@Arjun 

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I understood the question, But Isn’t there a conditional probablity. Because the system is declaring computers faulty if they are faulty.

And if this is conditional probablity then how can it be done using it?

Answer 1

Answer = option A

In image below the ticks show those branches where the result is declared as faulty.

So, required probability $=$ sum of probabilities of those two branches $= pq + (1-p)(1-q)$

Comments

nice amar

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Can we say here that

A : Assembly is faulty

B : Test gives correct result

A and B are independent

Hence P(A) = P(A $\cap$ B) + P(A $\cap$ B') = [P(A)*P(B) + P(A)*P(B')]

Answer 2

I think a is the correct option ..

prob = pq + (1-p)(1-q)  // in words it means

(it is faulty)(machine detects correctly) + (it is not faulty)(machine detects incorrectly)

Comments

missed + between pq and (1 - p)(1 -q)

Answer 3

A computer can be declared faulty only if--

it is faulty and testing process gives correct result(computer declared as faulty) OR it is not faulty and testing process gives incorrect result(that means computer declared as faulty).

$\implies\;pq+(1-p)(1-q)$

Answer 4

The probability of the computer being declared faulty is,
= Probability of testing process gives the correct result × Probability that computer is faulty + Probability of testing process giving incorrect result × Probability that computer is not faulty
= p × q + (1 - p) (1 - q)

Answer:A