11.02c, 11.02e - Sets, Venn diagrams and addition

11.02c, 11.02e - Sets, Venn diagrams and addition

If 18 students play football and 14 make music, must there be 32 different students? Under 11.02c you will describe sets and use Venn diagrams to keep overlapping groups separate. Under 11.02e you will calculate probabilities of “not”, “or” and “both” without counting anybody twice.

Describing sets

A set is a collection with a clear membership rule. For example, A={2,4,6}A=\{2,4,6\} lists three members, while “the even integers from 1 to 6” describes the same set. Order does not matter, and writing a member twice does not add a new member.

Set descriptions

Curly brackets enclose a set. A description such as A={x∣x is an even integer and 1≤x≤6}A=\{x\mid x\text{ is an even integer and }1\le x\le6\} means “the set of values xx such that the stated conditions hold”. The vertical bar means such that. The universal set is the complete collection being considered.

State the number type when it affects membership. Integers in an interval can be listed; all real numbers in that interval usually cannot be listed one by one.

Worked example 1: listing an integer set

Question. List D={x∣x is an integer and 1<x<7}D=\{x\mid x\text{ is an integer and }1<x<7\}.

Plan. Read both strict boundaries, then list each integer between them once.

1<x<7⇒x=2,3,4,5,6D={2,3,4,5,6}\begin{aligned} 1<x<7 &\quad\Rightarrow\quad x=2,3,4,5,6 \\ D &=\{2,3,4,5,6\} \end{aligned}

Answer. D={2,3,4,5,6}D=\{2,3,4,5,6\}.

Check. Every listed number is an integer satisfying both inequalities; the next integers, 11 and 77, fail a strict boundary.

Change only the upper inequality to allow equality. One new member enters.

Worked example 2: including a boundary

Question. List E={x∣x is an integer and 1<x≤7}E=\{x\mid x\text{ is an integer and }1<x\le7\}.

Plan. Keep the lower boundary strict and include the upper endpoint.

1<x≤7⇒x=2,3,4,5,6,7E={2,3,4,5,6,7}\begin{aligned} 1<x\le7 &\quad\Rightarrow\quad x=2,3,4,5,6,7 \\ E &=\{2,3,4,5,6,7\} \end{aligned}

Answer. E={2,3,4,5,6,7}E=\{2,3,4,5,6,7\}.

Check. The only change from DD is the permitted value 77; 11 is still excluded.

Sets can contain objects too: “students who cycle” is a membership rule. “The nicest students” needs an agreed criterion before it can classify people reliably. We will use the words “and”, “or” and “not” for relationships between sets; extra intersection, union and complement symbols are not needed.

Watch out

{1,2,2,3}\{1,2,2,3\} has three different members, not four. Also, {x∣1<x<3}\{x\mid1<x<3\} does not mean only {2}\{2\} unless xx is restricted to integers; for real xx, values such as 1.51.5 belong too.

Read both the domain and the boundaries.

Placing outcomes in Venn regions

Two properties can overlap. A Venn diagram puts all outcomes inside a rectangle representing the universal set, with a circle for each property. Someone with both properties goes in the overlap once, rather than appearing in two separate regions.

The four Venn regions

For sets HH and CC, the four separate regions are: HH only, both HH and CC, CC only, and neither. “HH or CC” includes the overlap. “Exactly one” excludes the overlap.

To construct the diagram, draw the rectangle and two labelled overlapping circles. Enter the overlap first, subtract it from each circle total, then use the overall total to find neither.

[DIAGRAM: asset_name: Filling two-circle Venn diagrams; asset_slug: venn-count-regions; caption: A circle total includes its overlap.; description: Two labelled Venn diagrams for forty students. Both have H total 22 and C total 17. The first has H-only 13, both 9, C-only 8 and neither 10. The second has H-only 17, both 5, C-only 12 and neither 6. All four regions lie inside each universal rectangle.]
A circle total includes its overlap.

Worked example 3: working from totals

Question. Of 4040 students, 2222 like hiking, 1717 like chess and 99 like both. Find all four Venn-region counts and the probability that a randomly chosen student likes exactly one activity.

Plan. Put the nine shared students in first, then subtract them from each group total.

H only=22−9=13C only=17−9=8neither=40−(13+9+8)=10p(exactly one)=13+840=2140\begin{aligned} H\text{ only} &=22-9=13 \\ C\text{ only} &=17-9=8 \\ \text{neither} &=40-(13+9+8)=10 \\ p(\text{exactly one}) &=\frac{13+8}{40}=\frac{21}{40} \end{aligned}

Answer. The regions are 13,9,8,1013,9,8,10; the probability is 2140\frac{21}{40}.

Check. The four regions total 4040, while the circle totals reconstruct 13+9=2213+9=22 and 9+8=179+8=17.

Keep the total and both circle totals unchanged, but reduce the overlap to five. Fewer shared members mean more different people inside the circles.

Worked example 4: changing the overlap

Question. There are again 4040 students, 2222 liking hiking and 1717 liking chess, but now 55 like both. Find the regions and the probability of exactly one activity.

Plan. Subtract the new overlap from each unchanged circle total.

H only=22−5=17C only=17−5=12neither=40−(17+5+12)=6p(exactly one)=17+1240=2940\begin{aligned} H\text{ only} &=22-5=17 \\ C\text{ only} &=17-5=12 \\ \text{neither} &=40-(17+5+12)=6 \\ p(\text{exactly one}) &=\frac{17+12}{40}=\frac{29}{40} \end{aligned}

Answer. The regions are 17,5,12,617,5,12,6; the probability is 2940\frac{29}{40}.

Check. 17+5+12+6=4017+5+12+6=40, and both original circle totals still hold.

A diagram can contain individual outcomes instead of counts. For numbers 11–66, “even” and “multiple of 3” place 66 in both, 2,42,4 in even only, 33 in multiple-of-3 only, and 1,51,5 outside both. The same four-region logic applies.

Watch out

Do not put 2222 into the hiking-only region when 2222 is the whole hiking total. If your four regions exceed the universal total, check for a duplicated overlap. Negative region counts signal inconsistent data or an error.

Give the new region counts in words; you do not need a drawing tool.

Events that do not happen

A parcel either arrives on time or does not. These two possibilities cover everything and cannot happen together. This makes “not” especially useful when an event has many possible outcomes.

Complement rule

For any event AA,

p(A)+p(not A)=1,p(not A)=1−p(A).p(A)+p(\text{not }A)=1, \qquad p(\text{not }A)=1-p(A).

An event and its complement have no shared outcomes and together cover the whole sample space. Not separately printed on the 2027 formula sheet: learn this rule.

The complement may be a collection of several named outcomes. “Not red” can include blue, green and any other colours, rather than meaning blue alone.

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