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math common language q

This question should be relatively easy according to the smc difficulty scale, but I didnt get it, and even looking at the solution and then working it out took me some time.

math.jpg
It seems to just list all the possibilities out and then picking the ones we care about, which seems time consuming - I was wondering if anyone has a faster/more efficient way (the ms has one but are there any others?).

Usually for probabilty qs I usually come in with venn/tree diagrams (which failed miserably with 6 people/4 languages), would anyone recommend any other tools, especially for smc? Like in the ms they just list them all out in a table and use that weird graph in the bottom right.

Help/advice greatly appreciated.
(edited 6 months ago)
Reply 1
Original post by mosaurlodon
This question should be relatively easy according to the smc difficulty scale, but I didnt get it, and even looking at the solution and then working it out took me some time.math.jpgIt seems to just list all the possibilities out and then picking the ones we care about, which seems time consuming - I was wondering if anyone has a faster/more efficient way (the ms has one but are there any others?).
Usually for probabilty qs I usually come in with venn/tree diagrams (which failed miserably with 6 people/4 languages), would anyone recommend any other tools, especially for smc? Like in the ms they just list them all out in a table and use that weird graph in the bottom right.
Help/advice greatly appreciated.

Either of those two methods table/intersecting graph would be ok, but there are 4 languages and there are 6 ways of choosing 2, so every pair is associated with a person and each language will occur in 3 pairs. So selecting person 1 (arbitrarily) does languages A and B

A is also done by 2 other people

B is also done by a different 2 other people

There are 5 other people so 4/5.

Or C&D is done by one other person so 4/5

Its just a basic 4C2. I guess you need to have the confidence that person 1 can be selected arbitrarily (without loss of generality) and think what happens about selecting person 2 and this works because every language pair is uniqely associated with a person.
(edited 6 months ago)
Reply 2
Ok I see... so its just a case of realising that its q14, there must be an easy way, and then spotting that if you pick one, then there are 4 other people out of the 5 who can speak your language, makes sense.

I really need to stack the practice up for these probability qs (and geometry yikes), but luckily my schools finally over so ill get some more time.

If you were sitting this q in the exam, would you list all the probabilties/outcomes out in a table? I always thought that was a bad idea since time, or would you do it by choosing 1 student and seeing what happens to the probabilties from there..?
Reply 3
Original post by mosaurlodon
Ok I see... so its just a case of realising that its q14, there must be an easy way, and then spotting that if you pick one, then there are 4 other people out of the 5 who can speak your language, makes sense.
I really need to stack the practice up for these probability qs (and geometry yikes), but luckily my schools finally over so ill get some more time.
If you were sitting this q in the exam, would you list all the probabilties/outcomes out in a table? I always thought that was a bad idea since time, or would you do it by choosing 1 student and seeing what happens to the probabilties from there..?

Its a basic counting question (dressed up as a probability) so once you spotted that
6 (students) = 4C2 (language pairs)
So its just a case of imagining (or writing down) the
AB, AC, AD, BC, BD, CD
pairs and choosing the first arbitrarily and simple reasoning about the others. Really, you should be able to do the reasoning in your head so consdier AB, then CD is the only one that shares nothing so 4/5.

Tbh, its one of those questions which a keen young kid could probably do (just listing the 6 pairs) and learning more probability theory (gcse ...) doesnt necessarily help/hinders the solution. Even just selecting the first person in the question and counting the others who did English or German would give 4/5.
(edited 6 months ago)
Reply 4
Original post by mosaurlodon
Ok I see... so its just a case of realising that its q14, there must be an easy way, and then spotting that if you pick one, then there are 4 other people out of the 5 who can speak your language, makes sense.
If you were sitting this q in the exam, would you list all the probabilties/outcomes out in a table? I always thought that was a bad idea since time, or would you do it by choosing 1 student and seeing what happens to the probabilties from there..?
It's worth noting that there doesn't have to be an easy way. You could have two students with the same language combo, and then all the "fancy" arguments fall apart. That isn't the case here, but you can't easily know it's not the case without doing something that's roughly "number of students"^2 operations.

In this kind of exam, if you see a method that's going to work and isn't going to take too long, I'd say it's generally best to go for it, or at least not to waste more than a few seconds looking for something better.
Reply 5
I dont know, usually if I see a method it tends to be a longwinded approach to a problem. Especially with geometry, because I tend to just brute force the calculations rather than looking at things that simplify like symmetry etc since I have terrible intuition. But I guess good intuition (where instantly seeing a method and then doing it to get the right answer works) just comes with practice.
Reply 6
Original post by mosaurlodon
I dont know, usually if I see a method it tends to be a longwinded approach to a problem. Especially with geometry, because I tend to just brute force the calculations rather than looking at things that simplify like symmetry etc since I have terrible intuition. But I guess good intuition (where instantly seeing a method and then doing it to get the right answer works) just comes with practice.

The thing is, any number of trivial and not totally obvious to spot changes would mean the "symmetry" arguments give the wrong answer. E.g. if you change it so that Mary speaks Spanish and French then you need a different argument.

So if you wanted to use symmetry, you'd need to verify that no language pairs are duplicated, at which point you've almost done the same work as the given solution does building the table. But, unlike that solution, you have "no backup plan". If there is a duplicated pair, you've basically got to find a completely different approach. The given solution method works regardless of how the languages are distributed.

As I understand it, you get less than 5 minutes per question. With those constraints, you sometimes need to pick a method (that you can see works and isn't too slow) and go for it. Filling in that table of 15 ticks/crosses is only going to take 30 seconds (drawing out the grid is going to take longer, but only a minute or so) so it's a perfectly viable solution.

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