Miss Loi waits patiently in the car, with sunglasses on.

12:00PM:

Target appears – spotted walking towards another car. Miss Loi’s eyes brightens. *Grips steering wheel*

12:02PM:

Aunty (i.e. The Target) takes an age to squeeze her groceries into car. Miss Loi getting impatient.

12:03PM:

Aunty completes squeezing of groceries into car. Aunty drives out of parking lot. Miss Loi takes off sunglasses and quickly zooms into vacated lot before the Ah Beng car behind beats her to it.

12:04PM:

After a thorough scan across the horizon, Miss Loi tears her parking coupon (like all good parking citizen should do), as shown below:

Miss Loi’s Parking Coupon – “scheduled” to start at 12:20PM

12:05:00PM:

Miss Loi hippity-hippity hops away from her car.

1:20PM:

Miss Loi hippity-hippity hops back to find this on her car:

Ninja Shuriken – recovered at 12:08PM (click to enlarge)

Bro! You wouldn’t believe how SUAY I was at the carpark just now! I tore my coupon to start at 12:20PM, left my car at 12:05PM, and I got summoned at 12:08PM! They employ ninjas as parking attendants these days is it?!!!!

Mr Loi replied in a calm voice:

Sigh … that is a ‘hot’ carpark with a parking attendant coming every hour. Please brush up your rusty A-Level Probability & Statistics and solve the following question to know your chances of getting caught, and try to lower your probability of getting caught the next time:

Assuming that the parking attendant visits the carpark once every hour, and that the probability of her turning up to check at any given minute is 1/60.

Miss Loi tears her coupon to commence at 12:20PM and left her car at 12:05:00PM.

Find the probability of the parking attendant first appearing at 12:08PM.

Find the probability of Miss Loi getting caught i.e. the parking attendant appearing at anytime between 12:05-12:19PM.

What is the latest commencing time she should tear in her coupon in order to have at most a 10% chance of getting caught by the parking attendant?

well they don't check the time via watches. Their "Summons-Issuing-Machine" have clock-sync capabilities so their time is always accurate (atomic clocks).

I don't have to be one of them to know. Just think about this logically (Maths is all about logic remember?)

Ticket issued to you, you argue/complain/appeal etc, and claim the time is incorrect. How do the authorities find a way to make sure this "incorrect time" argument is moot?

Answer - sync the clocks on the devices to atomic time. Nobody can argue with a clock that's only off by 1 second every few thousand years, and even so, the time is "off" only because Earth's rotation is slowing down, not because atoms don't vibrate on a consistent frequency 🙂

Miss Loi: The question asks for the probability of the parking attendant first appearing at 12:08PM, this means that she cannot appear at 12:05PM AND 12:06PM AND 12:07PM too!

2. 1/4 [15min * 1/60 = 15/60 = 1/4]

Miss Loi: This time the question asks for the probability of the parking attendant appearing anytime between 12:05-12:19PM. In addition, we're told that she visits the carpark once every hour. This means that she can only appear at 12:05PM OR 12:06PM OR 12:07PM OR ... OR 12:19PM.

Get the drift so far? 😉

3. 12:14PM? Coz 6min*1/60 = 6/60 = 1/10 = 10%

But then i think can only tear in 5min intervals. so i guess 12:15PM is more realistic ba. Miss Loi can enlighten anot?

Miss Loi: Your method of trial and improvement here is sound (which is what Miss Loi do in real life anyway), but you'll first need the correct expression from part 2.

1. P(PA first appearing at 12:08) = P(PA not appearing at 12:05-:07) * P(PA appearing at 12:08) = P(PA not appearing at 12:05) * P(PA not appearing at 12:06) * P(PA not appearing at 12:07) * P(PA appearing at 12:08) = 59/60 * 59/60 * 59/60 * 1/60 = 205,739/12,960,000 = 0.0158749 = 1.59%

2. P(appear once between 12:05 and 12:19) = P(appear once at 12:05) + ... + P(appear once at 12:19) = 15 * 1/60 * (59/60)^19 = 0.181658 = 18.17%

3. Since got 18.17% chance hor, then miss loi should tear at 12:15PM lor. Since 1/3 of 18.17% is 6+% 10%.

Thanks for attempting the question to determine how suay my sister was.

To simplify matters, we can first let:

p be the probability of the parking attendant (P.A.) turning up at any given minute

q (=1-p) be the probability of the P.A. NOT turning up at any given minute

X be the number of minutes it takes (starting from 12:05:00PM) for the P.A. to first turn up.

For Part 1, as Miss Loi has explained in Comment #12, we're in fact looking for P(X = 4). So you can't simply throw in 1/60 as the answer, since for the P.A. to first appear at 12:08PM, we also have to make sure that she DOESN'T appear at 12:05 AND 12:06 AND 12:07 AND DOES appear at 12:08.

So your revised workings in Comment #17 are spot-on in reflecting the expression: P(X=4) = q × q × q × p = q^{3}p = 0.0158

For Part 2, you're right to look for P(appear at 12:05) + … + P(appear at 12:19) as the probability of the P.A. appearing at anytime in the 15-minute range between 12:05-12:19PM.

However from what we've learned in Part 1, each of these probabilities i.e. P(appear at 12:05), P(appear at 12:06) etc. cannot be 1/60 but are actually:

P(appear at 12:05) = P(X=1) = q^{0}p P(appear at 12:06) = P(X=2) = q^{1}p P(appear at 12:07) = P(X=3) = q^{2}p . . . P(appear at 12:19) = P(X=15) = q^{14}p

Calculating each of these terms is time consuming, but when you sum them all up i.e.

You'll find that P(X ≤ 15) is in effect a Geometric Progression with first term a = p, common ratio r = q and number of terms n = 15.

Using , P(X ≤ 15) can be simplified to:

P(X ≤ 15) = = 0.223

At this point, we should be able to derive the general expression: P(X ≤ n) = 1-q^{n} ----- (1)

Errr ... Part 3 is actually the opposite of Part 2 - we're looking for the time range in which the probability of the P.A. turning up anytime within this time range is ≤ 10% i.e. P(X ≤ n) ≤ 0.1, where n is the number of minutes from 12:05:00PM.

So a more appropriate approach would be to make use of expression (1) from Part 2 as follows:

P(X ≤ n) ≤ 0.1 1-q^{n} ≤ 0.1 q^{n} ≥ 0.9 n lg(59/60) ≥ lg(0.9) n ≤ 6.268 ⇒ Max n = 6

So, the P.A. will have at most a 10% chance of appearing between 12:05:00PM to 12:11PM, which means that Miss Loi still has to tear at 12:15PM (∵ parking coupons are torn at 5-min intervals) Ha ha ha

P.S. Feel free to use the above expression to obtain other values of n in relation to different probability values. For e.g. you'll find that Miss Loi needs to tear at 12:10PM if she wishes to have at most a 5% chance of being caught - but I don't think she's genetically-programmed to do this.

Just to add further, the above question actually deals with something called a Geometric Distribution (I believe this is currently being taught in some schools but not all) which is used to describe situations that involve a number of independent (success or failure) trials, with the probability p of a successful outcome being the same for each trial.

In all geometric distributions, P(X ≤ n) = 1 - q^{n} (as shown above) P(X > n) = q^{n}

Clarion: Your name is on NParks' wanted list now for damaging the environment.

wj: Now you know that Miss Loi is an EXTREMELY patient tutor. Miss Loi fell asleep while waiting that night. While waiting, Miss Loi got abducted by aliens and they released her 12 hours later.

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.

About Miss Loi

Miss Loi is a full-time private tutor in Singapore specializing in O-Level Maths tuition. Her life’s calling is to eradicate the terrifying LMBFH Syndrome off the face of this planet. For over years she has been a savior to countless students … [read more]

Gratitudes

I was a F9, E8 person for maths … But after attending lessons for a month (4 sessions, 8 hours!), I jumped to A1! … [read more]

## 19 Comments

曜

日

Ninjas have been hired to specifically combat the likes of you. Cower in FEAR~!!!!!

曜

日

oh my...their skills seem to be getting better day by day! Go appeal. What if ninja's made-in-China watch is 10 minutes slower?

曜

日

well they don't check the time via watches. Their "Summons-Issuing-Machine" have clock-sync capabilities so their time is always accurate (atomic clocks).

Um yeah Singapore has an atomic clock too...

曜

日

Krisandro:*Grrrrr* ...Janice:Appeal? SeeFoxTwo's comment below yours: they're atomic-powered cyborgs!FoxTwo:How come you know so much?! Are you one of them??? You spotted my car and sent one of them to come at 12:08PM right? RIGHT??曜

日

I don't have to be one of them to know. Just think about this logically (Maths is all about logic remember?)

Ticket issued to you, you argue/complain/appeal etc, and claim the time is incorrect. How do the authorities find a way to make sure this "incorrect time" argument is moot?

Answer - sync the clocks on the devices to atomic time. Nobody can argue with a clock that's only off by 1 second every few thousand years, and even so, the time is "off" only because Earth's rotation is slowing down, not because atoms don't vibrate on a consistent frequency 🙂

曜

日

*raises hands*

Miss Loi, you mean you have never seen the Parking Pontianak before?

http://www.youtube.com/watch?v=QTTlk601I6E

曜

日

Appeal and say your clock faster than theirs.

曜

日

Wah, you put advance 15min? Power. The max I do is 10min. Depending on the "hotness" of the car park.

曜

日

Cockroach:You know that Miss Loi purposely waited till daytime before clicking on your link?!Folks, Miss Loi would like to share with you a helpful (albeit non-mathematical) "solution" left by a kind soul in her Facebook profile:

May it serve our nation well.

曜

日

So now you know the Parking Pontianaks knew all your little tricks eh?

曜

日

lol. ms loi. you are damn suay. ahahah.

曜

日

1. 1/60

Miss Loi:The question asks for the probability of the parking attendantfirstappearing at 12:08PM, this means that shecannotappear at 12:05PM AND 12:06PM AND 12:07PM too!2. 1/4 [15min * 1/60 = 15/60 = 1/4]

Miss Loi:This time the question asks for the probability of the parking attendant appearinganytimebetween 12:05-12:19PM. In addition, we're told that she visits the carparkonce every hour. This means that she canonlyappear at 12:05PM OR 12:06PM OR 12:07PM OR ... OR 12:19PM.Get the drift so far? 😉

3. 12:14PM? Coz 6min*1/60 = 6/60 = 1/10 = 10%

But then i think can only tear in 5min intervals. so i guess 12:15PM is more realistic ba. Miss Loi can enlighten anot?

Miss Loi:Your method of trial and improvement here is sound (which is what Miss Loi do in real life anyway), but you'll first need the correct expression from part 2.Hope you see this before Mr Loi comes along!

曜

日

you have facebook? :O add me! 😀

PS 我重出江湖了！xD

曜

日

Cockroach:Miss Loi recognizes that Pontianak!!!cy:辛灾乐祸!r0n:Miss Loi has left some comments beneath your workings. Hurry, hurry have a look before Mr Loi comes!Clarion:Miss Loi passed by a cave just now with many broken boulders scattered around its opening. Were you the one who burst out of it?曜

日

Eh how did you know? : p

曜

日

"11:59PM:Miss Loi waits patiently in the car, with sunglasses on.

12:00PM:Target appears - spotted walking towards another car. Miss Loi’s eyes brightens. *Grips steering wheel*"

Wow, you waited more than 12 hours for a parking lot?

曜

日

*furiously flips a lvl statistics textbooks*

1. P(PA first appearing at 12:08) = P(PA not appearing at 12:05-:07) * P(PA appearing at 12:08) = P(PA not appearing at 12:05) * P(PA not appearing at 12:06) * P(PA not appearing at 12:07) * P(PA appearing at 12:08) = 59/60 * 59/60 * 59/60 * 1/60 = 205,739/12,960,000 = 0.0158749 = 1.59%

2. P(appear once between 12:05 and 12:19) = P(appear once at 12:05) + ... + P(appear once at 12:19) = 15 * 1/60 * (59/60)^19 = 0.181658 = 18.17%

3. Since got 18.17% chance hor, then miss loi should tear at 12:15PM lor. Since 1/3 of 18.17% is 6+% 10%.

曜

日

Hello

r0n,Thanks for attempting the question

~~to determine how~~.suaymy sister wasTo simplify matters, we can first let:

pbe the probability of the parking attendant (P.A.) turning up at any given minuteq(=1-p) be the probability of the P.A. NOT turning up at any given minuteXbe the number of minutes it takes (starting from 12:05:00PM) for the P.A. tofirstturn up.For Part 1, as Miss Loi has explained in Comment #12, we're in fact looking for P(

X= 4). So you can't simply throw in 1/60 as the answer, since for the P.A. tofirstappear at 12:08PM, we also have to make sure that she DOESN'T appear at 12:05 AND 12:06 AND 12:07 AND DOES appear at 12:08.So your revised workings in Comment #17 are spot-on in reflecting the expression:

P(

X=4) =q×q×q×p=q^{3}p= 0.0158For Part 2, you're right to look for P(appear at 12:05) + … + P(appear at 12:19) as the probability of the P.A. appearing at anytime in the

15-minute rangebetween 12:05-12:19PM.However from what we've learned in Part 1, each of these probabilities i.e. P(appear at 12:05), P(appear at 12:06) etc. cannot be 1/60 but are actually:

P(appear at 12:05) = P(

X=1) =q^{0}pP(appear at 12:06) = P(

X=2) =q^{1}pP(appear at 12:07) = P(

X=3) =q^{2}p.

.

.

P(appear at 12:19) = P(

X=15) =q^{14}pCalculating each of these terms is time consuming, but when you sum them all up i.e.

P(

X≤ 15) =q^{0}p+q^{1}p+q^{2}p+ ... +q^{14}pYou'll find that P(

X≤ 15) is in effect a Geometric Progression with first terma=p, common ratior=qand number of termsn= 15.Using , P(

X≤ 15) can be simplified to:P(

X≤ 15) = = 0.223At this point, we should be able to derive the general expression:

P(

X≤n) = 1-q^{n}----- (1)Errr ... Part 3 is actually the opposite of Part 2 - we're looking for the time range in which the probability of the P.A. turning up anytime within this time range is ≤ 10% i.e. P(

X≤n) ≤ 0.1, wherenis the number of minutes from 12:05:00PM.So a more appropriate approach would be to make use of expression (1) from Part 2 as follows:

P(

X≤n) ≤ 0.11-

q^{n}≤ 0.1q^{n}≥ 0.9nlg(59/60) ≥ lg(0.9)n≤ 6.268 ⇒ Maxn= 6So, the P.A. will have at most a 10% chance of appearing between 12:05:00PM to 12:11PM, which means that Miss Loi still has to tear at 12:15PM (∵ parking coupons are torn at 5-min intervals) Ha ha ha

P.S. Feel free to use the above expression to obtain other values of

nin relation to different probability values. For e.g. you'll find that Miss Loi needs to tear at 12:10PM if she wishes to have at most a 5% chance of being caught - but I don't think she's genetically-programmed to do this.Just to add further, the above question actually deals with something called a Geometric Distribution (I believe this is currently being taught in some schools but not all) which is used to describe situations that involve a number of independent (success or failure) trials, with the probability

pof a successful outcome being the same for each trial.In all geometric distributions,

P(

X≤n) = 1 -q^{n}(as shown above)P(

X>n) =q^{n}曜

日

Clarion:Your name is on NParks' wanted list now for damaging the environment.wj:~~Now you know that Miss Loi is an EXTREMELY patient tutor.~~~~Miss Loi fell asleep while waiting that night.~~~~While waiting, Miss Loi got abducted by aliens and they released her 12 hours later.~~Alright fine! It was a typo!