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

The key idea is conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, and careful use of "fails alone" to isolate events.

Step 1: Define events.

Let $F_A$ = "A fails", $F_B$ = "B fails". Given:

$P(F_A) = 0.2$, $P(F_A \cap F_B) = 0.15$, and $P(\text{B fails alone}) = P(F_B \cap F_A^c) = 0.15$.

Step 2: Find $P(F_B)$.

Since $F_B = (F_B \cap F_A) \cup (F_B \cap F_A^c)$, and these are disjoint,

$P(F_B) = P(F_A \cap F_B) + P(F_B \cap F_A^c) = 0.15 + 0.15 = 0.3$.

Step 3: Compute (i). …

We are given three probabilities:

• $P(A \text{ fails}) = 0.2$
• $P(B \text{ fails alone}) = 0.15$
• $P(A \text{ and } B \text{ fail}) = 0.15$

The phrase "B fails alone" means B fails and A does not fail. In set notation: $P(B \text{ fails} \cap A \text{ does not fail}) = 0.15$.

Similarly, "A fails alone" means $P(A \text{ fails} \cap B \text{ does not fail})$, which we need to find in part (ii).

Step-by-step reasoning

1. Understand the events and notation

Let:

• $A$ = event that subsystem A fails
• $B$ = event that subsystem B fails

We know:

• $P(A) = 0.2$
• $P(A \cap B) = 0.15$
• $P(B \cap A^c) = 0.15$ (B fails alone)

We want:

• (i) $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$
• (ii) $P(A \cap B^c)$ (A fails alone)

2. Find $P(B)$ first

The event "B fails" can happen in two mutually exclusive ways:

• B fails and A fails (joint failure)
• B fails and A does not fail (B alone)

So:

Substitute the known values:

3. Compute $P(A \mid B)$

Using the definition:

$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.15}{0.3} = 0.5$ …

Method: Decomposing events with "alone", joint, and conditional relationships

Use this when a problem mixes phrases like "$A$ fails", "$B$ fails alone", and "$A$ and $B$ fail", and asks for a conditional or an "alone" probability.

Steps

Step 1: Translate every phrase into set language.

"$X$ and $Y$ both fail" is $P(X \cap Y)$; "$X$ fails alone" is $P(X \cap Y^{c})$ — $X$ fails while the other does not. Keeping this distinction straight is the whole battle.

Step 2: Recover a marginal probability by splitting on the other event.

Any event splits into its overlap with $Y$ and with $Y^{c}$:

$P(X) = P(X \cap Y) + P(X \cap Y^{c}).$ …

Common Mistakes

Mistake 1: Taking "B fails alone" $=0.15$ as $P(B)$.

Why it's wrong: "$B$ alone" excludes the case where $A$ also fails, so it is only part of $P(B)$. Correct approach: $P(B)=P(A\cap B)+P(B\text{ alone})=0.15+0.15=0.3$.

Mistake 2: Computing $P(A\mid B)$ with the wrong denominator.

Why it's wrong: the denominator must be the full $P(B)=0.3$, not $0.15$. Correct approach: $P(A\mid B)=\dfrac{0.15}{0.3}=0.5$. …