Arjun Suresh (talk | contribs) |
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| − | Consider a company that assembles computers. The probability of a faulty | + | 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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| − | + | '''(A) $pq + (1 - p)(1 - q)$''' | |
| − | + | (B) $(1 - q)p$ | |
| − | + | (C) $(1 - q)p$ | |
| − | + | (D) $pq$ | |
==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
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) $pq$
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.
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) $pq$
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.
GATECSE