Dear Diary,

I've flipped through all your pages from front to back (and vice versa) but, for the life of me, couldn't find any mention of me being in an "all girls school".

]]>Dear Diary,

After enduring the most brutal dressing down ever dished out to me in my life, we were finally released from the common room.

Miss Lim will be speaking to our parents soon, but the most hurtful thing to me now is Miss Lim's suspicion that all my previous good results had been achieved through cheating.

Nonetheless I found some small solace in that I'm now walking beside HIM, and took the opportunity to ask him if our Valentine's date tonight is still on, given the chance that Miss Lim might not be calling our parents until tomorrow.

To my utter horror, HE snapped at me

YOU KNOW WHAT? I TOTALLY REGRET ASKING YOU TO HELP ME IN THE TEST!

IT'S ALL YOUR FAULT!

I told him that our chances of success was low, given that the overall probability of getting caught was 0.609 (3 sig. fig.)

PLEASE LAH! OF ALL THE GIRLS I'VE COLLABORATED WITH, YOU ARE THE ONLY ONE WHO HAVE FAILED ME! ALL OF THEM COULD DO IT MORE DISCREETLY THAN YOU!

I HAD NO CHOICE BUT TO TURN TO YOU BECAUSE THE ANSWERS GIVEN BY THOSE BIMBOS WERE MOSTLY WRONG SO I STILL FAILED!

I THOUGHT YOU BEING THE BEST IN MATH CAN SAVE ME BUT YOU TURNED OUT TO BE THE MOST USELESS OF ALL!

With that, he turned his back on me and stormed off.

*cues for camera to zoom away from my face with a rapidly receding background*

It hurt like a thousand roses had just wilted in my heart. I never knew how long I stood on that spot since his departure.

I've failed HIM. And now Miss Lim thinks I've been a cheater my whole life. And I've just used up this month's pocket money on that dress from B B Fashion.

I have never felt so miserable all my life. I …

*CUT!!!*

**At this point, Miss Loi slammed shut the pages of the diary following a sudden, chilling realization that HE happens to bear a vague resemblance to someone whom she had met not long ago.**

Dear Diary,

I now stand trembling (with HIM by my side) before Miss Lim's desk in the teachers' common room, with Miss Lim's notorious death gaze burning a searing path of guilt through my soul.

Looking back at what happened an hour ago ...

Deriving from the Probabilities lesson taught last week, I understand that the possible paths Miss Loi could take could be represented by a Probability Tree Diagram.

However as we are only interested in those paths that lead to my demise, it's unnecessary to sketch the entire tree but instead recognize that each junction in the classroom corresponds to a node in the tree and the possible directions to take at each junction are like the 'branches' from each node i.e. they are *mutually exclusive events* since Miss Lim does not possess the 分身术 ability to go straight, left and right at the same time *shudders*

Hence using the P(not `E`) = 1 − P(`E`) property described here, we can derive all the probabilities of the directions Miss Lim takes at each junction from the info given in the question, based on the number of possible choices she has at each junction:

P.S. It *may* not be obvious to some that whenever there's only **one** possible choice, P(this choice) = 1 *duh*

So with that settled, I can proceed to plot (with some prompting from a voice from the future) all the possible paths Miss Lim could take to reach my 'danger zone' i.e. Miss Lim's Six Paths to My Damnation.

Since Miss Lim's decision to go straight/turn left/right at each junction is not affected by her decisions at any previous junction, they are all *independent events* so the probability of each path is computed simply by multiplying her probabilities at all junctions along the way.

Now, since I only have one attempt to pass my answers to HIM, I'll get caught in this attempt only if Miss Lim takes Path 1 OR Path 2 OR Path 3 OR Path 4 OR Path 5 OR Path 6 i.e.

P(caught in one attempt)

= 0.144 + 0.0576 + 0.02592 + 0.0576 + 0.064 + 0.02592

= 0.37504

After going through such a long working, some of my mathematically-weaker classmates may be tempted to declare this as the final answer. However there are *two* attempts (i.e. I pass to HIM and HE passes it back to me) that needs to executed which affects the overall probability of being caught.

At this junction, I would instantly recognize this as a classic 'two-attempt' probability question that is commonly encountered throughout the TYS with the following probability tree:

As aptly summarised by the freaky voice from the future,

This question made me deliberated between multiplying or squaring the original probability value that I got, due to its "special nature". You see, had the students been caught on the first try of cheating, there would not be a second attempt. A student may, on first sight, simply just multiplied P(caught) by 2. If so, that student have well fallen into the "trap" of this question.

So instead of blindly × 2 or squaring P(caught) like some of my (forever failing) classmates would likely have done, it's important to do some split-second common-sensical analysis and convince yourself it tallies with what's described in the probability tree i.e.

Overall probability of being caught

= P(caught in 1st attempt) OR P(caught in 2nd attempt)

= P(caught in 1st attempt) + P(success in 1st attempt) × P(caught in 2nd attempt)

= 0.37504 + (1 − 0.37504)(0.37504)

≈ 0.609 (3 sig. fig.)

Since there's more than 60% chance of being caught, my mathematical subconsciousness implored me not to do it, but sadly love is blind ... and that's why I'm trembling in front of Miss Lim now 🙁

]]>(0.6 * 0.6 * 0.4) +

(0.6 * 0.4 * 0.4 * 0.6) +

(0.4 * 0.6 * 0.4 * 0.6) +

(0.4 * 0.4 * 0.4) +

(0.6 * 0.4 * 0.3 * 0.6 * 0.6) +

(0.4 * 0.6 * 0.3 *0.6 * 0.6)

= 0.37504 let the answer be V (for valentine's :P).

After my original answer (guess you know who i am now), I went back to recalculate the answer. I was careless in some junctions, due to the many many conditions which made me a little confused, as well as the fact that i was doing it mentally (i know, no excuse). That was my first mistake.

My second, is one I think allows some thought. This question made me deliberated between multiplying or squaring the original V-value that I got, due to its "special nature". You see, had the students been caught on the first try of cheating, there would not be a second attempt. A student may, on first sight, simply just multiplied V by 2. If so, that student have well fallen into the "trap" of this question. I drew out a probability tree diagram to better "see" the situation. Because I am lazy and I don't know how to attach images, I shall allow Ms Loi to do it for me. But, the gist is, to get the final probability of whether the students will get caught by Ms Lim, one must add V to the to the product of (1-V) (1st attempt successful) and V (2nd attempt failure).

P(getting caught during test)

= V + V(1-V) = 0.609424998 = 0.609 (3sf)

Therefore, I present the answer, 0.609. Or at least what I think is the right answer.

PS: If I'm right, please post the aftermath of the cheating please :3

]]>Someone on Facebook managed to obtain the 6 paths for Miss Lim to catch the ~~young Miss Loi~~ cheating students, and got an answer that's very close to Miss Loi's.

To investigate where the answers differ, here's Miss Loi's plot of one of the paths (let's call it Path 1):

Does this coincide with one of yours? Pay attention to the probability at each junction - they depend on the number of choices Miss Lim has at the junction, based on the restrictions described in the question.

P.S. The diagram has been flipped from top to bottom for better visualization from Miss Lim's point of view.

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