1. Define the events

Let:

• $A$ = motorcycle manufactured at plant $A$
• $B$ = motorcycle manufactured at plant $B$
• $S$ = motorcycle is of standard quality

Given:

• $P(A)=0.6$
• $P(B)=0.4$
• $P(S\mid A)=0.8$
• $P(S\mid B)=0.9$

We need to find: $p=P(B\mid S)$

2. Use Bayes' theorem

$P(B\mid S)=\frac{P(B\cap S)}{P(S)}=\frac{P(B)P(S\mid B)}{P(A)P(S\mid A)+P(B)P(S\mid B)}$

Substitute the values: $p=\frac{0.4\times 0.9}{0.6\times 0.8+0.4\times 0.9}$

3. Simplify

Numerator: $0.4\times 0.9=0.36$

Denominator: $0.6\times 0.8+0.4\times 0.9=0.48+0.36=0.84$

So, $p=\frac{0.36}{0.84}=\frac{36}{84}=\frac{3}{7}$

4. Find $126p$

$126p=126\times \frac{3}{7}=18\times 3=54$

5. Match with the options

$54$ corresponds to Option A.

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So, the answer agrees with the stored correct answer.