Difference between revisions of "GATE2010 q26"

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Consider a company that assembles computers. The probability of a faulty
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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?
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assembly of any computer is p. The company therefore subjects each
 
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computer to a testing process. This testing process gives the correct result for
 
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any computer with a probability of q. What is the probability of a computer
 
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being declared faulty?
 
 
 
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'''(A) pq + (1 - p)(1 - q)'''
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'''(A) $pq + (1 - p)(1 - q)$'''
 
 
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(B) (1 - q)p
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(B) $(1 - q)p$
 
 
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(C) <(1 - q)p
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(C) $(1 - q)p$
 
 
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(D) pq
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(D) $pq$
  
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
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P(declared faulty) = P(actually faulty)*P(declared faulty|actually faulty) +  
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P(declared faulty)  
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P(not faulty)*P(declared faulty|not faulty) = p*q + (1-p)*(1-q).
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= P(actually faulty)*P(declared faulty|actually faulty) + P(not faulty)*P(declared faulty|not faulty)  
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<br>
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= $p*q + (1-p)*(1-q)$
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So option <b>(A)</b> is correct.
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So, option <b>(A)</b> is correct.
  
 
{{Template:FBD}}
 
{{Template:FBD}}
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[[Category:Probability and Combinatorics]]
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[[Category: GATE2010]]
 
[[Category: GATE2010]]
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[[Category: Previous year GATE questions]]
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[[Category: Probability questions from GATE]]

Latest revision as of 11:53, 15 July 2014

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) $(1 - q)p$

(D) $pq$

Solution by Happy Mittal

P(declared faulty) 
= P(actually faulty)*P(declared faulty|actually faulty) +  P(not faulty)*P(declared faulty|not faulty) 
= $p*q + (1-p)*(1-q)$

So, option (A) is correct.




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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 - q)p

(D) pq

Solution by Happy Mittal[edit]

P(declared faulty) = P(actually faulty)*P(declared faulty|actually faulty) + P(not faulty)*P(declared faulty|not faulty) = p*q + (1-p)*(1-q).
So option (A) is correct.




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