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(B) (1 - q)p
 
(B) (1 - q)p
 
 
(C) <(1 - q)p
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(C) (1 - q)p
 
 
 
(D) pq
 
(D) pq
  
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
P(declared faulty) = P(actually faulty)*P(declared faulty|actually faulty) +  
+
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).
 
P(not faulty)*P(declared faulty|not faulty) = p*q + (1-p)*(1-q).
<br>
+
So option <b>(A)</b> is correct.
+
So, option <b>(A)</b> is correct.
  
 
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{{Template:FBD}}

Revision as of 22:08, 14 April 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.




blog comments powered by Disqus