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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability3 Marks
Question
An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0.2
P(B fails alone) = 0.15
P(A and B fail) = 0.15
Evaluate the following probabilities:
  1. P(A fails|B has failed)
  2. P(A fails alone)
✓

Answer

Consider the following events: E : A fails F : B fails. Now P(A fails) = 0.2 ⇒ P(E) = 0.2 And P(A and B fails) = 0.15 $\Rightarrow\ \text{P}(\text{E}\cap\text{F})= 0.15$ Also P(B fails alone) = 0.15 $\Rightarrow\ \text{P}(\overrightarrow{\text{E}}\cap\text{F})=0.15$ $\Rightarrow\ \text{P}(\text{F})-\text{P}({\text{E}}\cap\text{F})=0.15$ $\Rightarrow\ \text{P}(\text{F})-\text{P}({\text{E}}\cap\text{F})+0.15$ ⇒ P(F) = 0.15 + 0.15 = 0.30
  1. P(Afails| B has failed) $=\text{P}(\text{E|F})=\frac{\text{P}({\text{E}}\cap\text{F})}{\text{P}(\text{F})}= \frac{0.15}{0.30}=\frac{15}{30}=\frac{1}{2}$
  2. P(A fails alone) $=\text{P}(\text{E}\cap\overrightarrow{\text{F}})=\text{P}(\text{E})-\text{P}(\text{E}\cap\text{F})$
= 0.2 - 0.15 = 0.05

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