E9.1, E9.3 - Probability from Venn diagrams (Extended)

E9.1, E9.3 - Probability from Venn diagrams (Extended)

If two club lists each include the same student, can you add their sizes to find the chance of selecting a club member? E9.1 and E9.3 use probability notation with Venn diagrams, account for overlapping events and distinguish independent events from mutually exclusive ones.

Intersections and unions

A Venn diagram groups possible outcomes into regions. When one person is selected at random, each person has the same chance, so the probability of a region is its count divided by the whole group count.

Probability of intersections and unions

P(A∩B)P(A\cap B) means the probability of being in both A and B. P(A∪B)P(A\cup B) means the probability of being in A or B or both. The prime in P(A′)P(A') means not in A.

Use the set boundary to select regions before calculating. An intersection is not necessarily just one region when a third set is present: A∩BA\cap B includes any outcomes also in C.

[DIAGRAM: asset_name: Probability regions in two sets; asset_slug: two_set_probability; caption: The union includes three regions, while the intersection is the overlap.; description: A universal rectangle containing twenty-four people has overlapping circles A and B. A-only contains eight, the overlap five, B-only seven and outside both four. A caption identifies twenty people in the union.]
The union includes three regions, while the intersection is the overlap.

Worked example 1: select the regions

Question. There are 8 people in A only, 5 in both A and B, 7 in B only and 4 in neither. One person is selected at random. Find P(A∩B)P(A\cap B) and P(A∪B)P(A\cup B).

Plan. Use five people for both and count every inside region once for the union.

Total=8+5+7+4=24P(A∩B)=524P(A∪B)=8+5+724=2024=56\begin{aligned} \text{Total}&=8+5+7+4=24\\ P(A\cap B)&=\frac5{24}\\ P(A\cup B)&=\frac{8+5+7}{24}=\frac{20}{24}=\frac56 \end{aligned}

Answer. P(A∩B)=5/24P(A\cap B)=5/24, P(A∪B)=5/6P(A\cup B)=5/6.

Check. The complement of the union has four people. Thus P((A∪B)′)=4/24=1/6P((A\cup B)')=4/24=1/6, and 1−1/6=5/61-1/6=5/6.

Now add a third set. Keep the full pairwise intersection, including its central triple overlap.

[DIAGRAM: asset_name: Three-set probability regions; asset_slug: three_set_probability; caption: A intersection B contains 3 plus 4, and A union B includes every region inside either of those circles.; description: A three-circle Venn diagram inside a universal rectangle of thirty people has A-only six, B-only five, C-only four, A-and-B-only three, A-and-C-only two, B-and-C-only one, all three four, and outside all sets five. All eight regions are labelled clearly.]
A intersection B contains 3 plus 4, and A union B includes every region inside either of those circles.

Worked example 2: an intersection inside three sets

Question. A group of 30 has these disjoint region counts: A only 6, B only 5, C only 4; A and B only 3, A and C only 2, B and C only 1; all three 4; none 5. A person is selected at random. Find P(A∩B)P(A\cap B) and P(A∪B)P(A\cup B).

Plan. Include the centre in both events, but count it only once in each probability.

P(A∩B)=3+430=730P(A∪B)=6+5+3+2+1+430=2130=710\begin{aligned} P(A\cap B)&=\frac{3+4}{30}=\frac7{30}\\ P(A\cup B)&=\frac{6+5+3+2+1+4}{30}=\frac{21}{30}=\frac7{10} \end{aligned}

Answer. 7/307/30 and 7/107/10, respectively.

Check. Outside A and B are the 4 in C only and the 5 in none. Their probability is 9/309/30, leaving 1−9/30=21/301-9/30=21/30 in the union. The pairwise intersection has 3 outside C and 4 inside C.

If a diagram contains probabilities rather than counts, add the selected region probabilities directly. If it contains individual outcomes, count those outcomes first. Its labels tell you which representation is being used.

Watch out

“A and B” includes the centre of a three-set diagram. “A and B only” excludes C. The word “only” changes the selected region.

Use every relevant region, including the centre.

Overlapping events

Adding the whole of A to the whole of B counts their overlap twice. Subtract one copy of the overlap to leave each person or outcome counted once.

Union with an overlap

P(A∪B)=P(A)+P(B)−P(A∩B).P(A\cup B)=P(A)+P(B)-P(A\cap B).

The same counting relationship is n(A∪B)=n(A)+n(B)−n(A∩B)n(A\cup B)=n(A)+n(B)-n(A\cap B).

Not in the supplied formula list: learn it.

The correction is one overlap, not two: the addition included two copies and the union needs one. Drawing the disjoint regions gives an alternative to remembering a formula.

Worked example 3: find a union

Question. In a group of 40, 22 people belong to A, 18 to B and 9 to both. One person is selected at random. Find P(A∪B)P(A\cup B).

Plan. Subtract one overlap from the sum of the two set totals.

n(A∪B)=22+18−9=31P(A∪B)=3140\begin{aligned} n(A\cup B)&=22+18-9=31\\ P(A\cup B)&=\frac{31}{40} \end{aligned}

Answer. 31/4031/40.

Check. A only has 22−9=1322-9=13, B only has 18−9=918-9=9, and both has 9. These disjoint counts total 13+9+9=3113+9+9=31.

The next question gives the union indirectly through its complement. Reverse the same relationship to recover the overlap.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account