n=40

Today in 2C MAT we came across that old chestnut, the Venn diagram with the missing value in the intersection with a number in A, B and the outside region.

In many cases the easiest way is to use a guess and check approach and a lot of the time the answer will fall out by substituting into the intersection and revising your result based on the values

A union B + the outside region = n.

n=40

Another approach is to name the segments and solve a series of equations:

a = 20-b

c = 30-b

a + b + c + 5 = 40

By substitution (20-b) + b + (30 - b) + 5 = 40

Therefore b=15

Once the intersection(b) is known, finding "A only"(a) and "B only"(b) is trivial.

I was asked the question "why teach this technique?" and my response was that it was not formally taught, it was a logical answer for a question given. We have some unknowns, we have some equations, why not solve for them? This sort of problem solving "setting up of equations" technique is common in optimisation and linear programming - why not use it in a probability setting?

I remember a particular student that was renowned for having solutions of this nature where his answers always deviated from the answer key and he had the right answer (or was on the right track) more often than not. We still call intuitive answers like this after "that" student as they forced the marker to find the underlying logic rather than application of a given method (if that student is reading this - get offline and study for your uni courses, scallywag!)

Anyhow, a third and more common approach is to rearrange the property:

A U B = A + B - A intersection B

By rearranging the equation

A intersection B = A + B - AUB

Since we know that:

AUB = U - (the outside region)

to find AUB is fairly simple:

AUB = 40-5

= 35

Therefore:

A intersection B = 20 + 30 - 35

= 15 (as before)

This approach does have the advantage that you can talk about the intersection being counted twice when the union is calculated by adding A + B where A and B aren't mutually exclusive.

I can't really see how this problem could be classed complex given the second method exists. Perhaps, if combined with a wordy explanation, a question of this sort could be made complex but to my mind that would defeat the purpose of the syllabus points in defining complexity. After all, why should something be classed a "complex question" if the only reason was that the question was worded to be understood by students with strong English comprehension?

teach me!!!

ReplyDeletethank you.

ReplyDeletewill come in handy for todays exam.

Thank you!! Legit jist got it!!

ReplyDeleteU=100

ReplyDeleteS=70

F=50

I=30

All have At least two elements have 44 or all intersects sum up to 44 find S n F n I=?

hi

ReplyDelete