The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+12 votes
1.9k views

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?

  1. $pq + (1 - p)(1 - q)$
  2. $(1 - q)p$
  3. $(1 - p)q$
  4. $pq$
asked in Probability by Boss (18.3k points)
edited by | 1.9k views
+1

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

@Arjun 

3 Answers

+28 votes
Best answer

answer = option A

in image below the ticks shows those branch where the result is declared as faulty.

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

answered by Boss (31.1k points)
edited by
0
nice amar
0
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')]
+14 votes
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)
answered by Boss (14.4k points)
edited by
0
missed + between pq and (1 - p)(1 -q)
0 votes

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).

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

answered by Active (4.7k points)
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
47,922 questions
52,324 answers
182,350 comments
67,780 users