एक सॉफ्टवेयर इंजीनियर एक त्रुटि X को सही ढंग से डीबग करेगा, इसकी संभावना 80% है। सही डिबगिंग के बाद त्रुटि X वाला सॉफ़्टवेयर क्रैश होने की संभावना 30% है। गलत डीबगिंग द्वारा सॉफ़्टवेयर के क्रैश होने की संभावना 70% है। X त्रुटि वाला सॉफ़्टवेयर क्रैश हो गया। इसकी त्रुटि के सही ढंग से डिबग होने की प्रायिकता है:

This question was previously asked in
SSC CGL Tier-II ( JSO ) 2019 Official Paper ( Held On : 17 Nov 2020 )
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  1. \(\dfrac{7}{12}\)
  2. \(​​\dfrac{7}{19}\)
  3. \(​​\dfrac{6}{19}\)
  4. \(​​\dfrac{12}{19}\)

Answer (Detailed Solution Below)

Option 4 : \(​​\dfrac{12}{19}\)
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Detailed Solution

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दिया गया है

P(A I B) = 30% = 0.3

P(A) = 80% = 0.8

P(B I A’) = 70% = 0.7

P(A’) = 100% - 80% = 2-0% = 0.2

सूत्र

P(A I B) = P(A I B). P(A)/P(B) = [P(A I B). P(A)]/[P(A I B). P(A) + P(B I A’) P(A’)]

गणना

P(A I B) = (0.3 × 0.8)/(0.3 × 0.8 + 0.7 + 0.2)

⇒ 0.24/(0.24 + 0.14)

⇒ 0.24/0.38

इसके सही ढंग से त्रुटिपूर्ण होने की प्रायिकता 12/19 है​ 
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