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Q9(i):
An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$, $P(B \text{ fails alone}) = 0.15$, $P(A \text{ and } B \text{ fail}) = 0.15$. Evaluate the following probabilities (i) $P(A \text{ fails|B has failed})$

Solution :

Given Variables & Initial Setup

Based on the problem statement, we define the following events:

  • Let $A$ be the event that subsystem A fails. Thus, $P(A) = 0.2$.
  • Let $B$ be the event that subsystem B fails.
  • The probability that B fails alone is given as $P(B \cap A^c) = 0.15$.
  • The probability that both A and B fail is given as $P(A \cap B) = 0.15$.

Step 1: Determine the total probability of event B

The total probability of event $B$ failing is the sum of the probability that B fails while A fails, and the probability that B fails while A does not fail.

$P(B) = P(A \cap B) + P(B \cap A^c)$ [By the Law of Total Probability]

$P(B) = 0.15 + 0.15 = 0.30$

Step 2: Applying the Definition of Conditional Probability

We are required to find $P(A \text{ fails} | B \text{ has failed})$, which is denoted as $P(A|B)$.

The formula for conditional probability is:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ [Per the Axiom of Conditional Probability]

Step 3: Substitution and Calculation

Substitute the values derived in the previous steps:

$P(A|B) = \frac{0.15}{0.30}$

$P(A|B) = \frac{15}{30} = 0.5$

Visual Representation of Event Spaces

A B 0.15 0.05 0.15 Sample Space

Final Solution: $P(A|B) = 0.5$


More Questions from Class 12 Mathematics Probability Miscellaneous Exercise on Chapter 13


CBSE Solutions for Class 12 Mathematics Probability


Chapters in CBSE - Class 12 Mathematics


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