Miss Loi: Yup this is simple.

Since it's been stated that there's no chance of running into more than one aunty during her visit, and that each Aunty can ask her only one question when our Student runs into her,

• ⇒ meeting Aunty A and meeting Aunty B and meeting Aunty C are mutually exclusive events.
• ⇒ the questions asked by each Aunty are also mutually exclusive events.

i.e. Mutually exclusive events = when Event A occurs, then there's absolutely ZERO chance of Event B occurring.

So the only possibilities of our Student being asked about her O Level results,

P(asked O Level results) =
P(Run into Aunty A) AND P(Aunty A asks O Level results) OR
P(Run into Aunty B) AND P(Aunty B asks O Level results) OR
P(Run into Aunty C) AND P(Aunty C asks O Level results)

N.B. EMaths Probability calculations are usually pretty very straightforward. Often, an evil examiner's only recourse to 'challenge' you is via 'confusing' you with phrases like "% chance of running into Aunty B" - which you should instantly laugh into his face and say "Nah just say lah!"

Yup - continuing with the same reasoning in Part 1,

P(asked O Level results OR JC/Poly choice) =
P(asked O Level results) OR P(asked JC/Poly choice)
P(asked O Level results) OR
P(Run into Aunty A) AND P(Aunty A asks JC/Poly choice) OR
P(Run into Aunty B) AND P(Aunty B asks JC/Poly choice) OR
P(Run into Aunty C) AND P(Aunty C asks JC/Poly choice)

N.B. Different workings but same answer!

*Deng deng deng deng* ... unfortunately in a longish story-telling question like this, the euphoria one experiences from the relatively-straightforward earlier parts usually precedes something more sinister in the later parts. In this instance, there can be a limit of how far one can go, in terms of accuracy, when attempting Probability questions without the aid of a diagram:

While Miss Loi said earlier that EMaths Probability calculations are straightforward, the real challenge usually lies in the extraction of information from the question, as well as deducing what the question doesn't tell you. In this case, it's often about

Looking at the RED portion of the Probability Tree Diagram above, besides the case of not running into any Aunty, it's obvious there are another two cases when the Student can escape questioning i.e. when Aunty A or Aunty B keeps quiet after meeting the Student.

So P(escape questioning/scrutinizing) =
P(not running into any Aunty) OR
P(Run into Aunty A) AND P(Aunty A keeps quiet) OR
P(Run into Aunty B) AND P(Aunty B keeps quiet)

Hence a quick sketch of a Probability Tree Diagram (even when not asked) in your exam can be invaluable in helping you to see at a glance ALL possible outcomes and ensure that you don't miss out any. Also, in the event of a chorus of complaints/protests in the likes of "the question not phrased properly lah", "question not clear lah" ... it's worth noting that the sum of all probabilities branching out from each node must be 1.

So alarm bells should be ringing when you check that the given in the question for Aunty A, and the given for Aunty A, aren't equal to 1. Aunty C is really evil and mean as ⇒ she'll die die question/scrutinize our Student if she unfortunately runs into her

You've correctly covered all the cases for P(escaping any questioning/scrutinizing in ≥ 2 visits) but unfortunately your final answer is affected by the value of P(escape questioning/scrutinizing) you've obtained in Part 1.

From Part 1, we've obtained

P(she'll be asked/scrutinized on a visit) =

Since she can only either escape or fail to escape being asked/scrutinized on each visit (i.e. these are the only two possible outcomes for each visit),

⇒ P(she'll NOT be asked/scrutinized on a visit) =

To make things a bit clearer, here's a second Probability Tree Diagram for the three house visits in Part 2:

From the diagram, we can easily see that for P(escaping any questioning/scrutinizing in ≥2 visits), there're only the following 'paths' that she can take:

1. Escape -> Escape -> Escape
2. Escape -> Escape -> No Escape
3. Escape -> No Escape -> Escape
4. No Escape -> Escape -> Escape

∴ P(escaping any questioning/scrutinizing in ≥2 visits)

Following what's described above, P(having a happy Chinese New Year) should then be