Arjun Suresh (talk | contribs) |
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 | + | assembly of any computer is $p$. The company therefore subjects each |
computer to a testing process. This testing process gives the correct result for | 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 | + | any computer with a probability of $q$. What is the probability of a computer |
being declared faulty? | being declared faulty? | ||
| − | '''(A) pq + (1 - p)(1 - q)''' | + | '''(A) $pq + (1 - p)(1 - q)$''' |
| − | (B) (1 - q)p | + | (B) $(1 - q)p$ |
| − | (C) (1 - q)p | + | (C) $(1 - q)p$ |
| − | (D) pq | + | (D) $pq$ |
==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
P(declared faulty) | P(declared faulty) | ||
= P(actually faulty)*P(declared faulty|actually faulty) + P(not faulty)*P(declared faulty|not faulty) | = P(actually faulty)*P(declared faulty|actually faulty) + P(not faulty)*P(declared faulty|not faulty) | ||
| − | = p*q + (1-p)*(1-q) | + | = $p*q + (1-p)*(1-q)$ |
So, option <b>(A)</b> is correct. | So, option <b>(A)</b> is correct. | ||
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[[Category: GATE2010]] | [[Category: GATE2010]] | ||
[[Category: Probability questions]] | [[Category: Probability questions]] | ||
| − | [[Category:Mathematics | + | [[Category:Mathematics questions ]] |
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