Probability with Venn Diagrams

Worked example

1. 1

   $\text{total}=14+8+11+17=50$ — Add all four regions. Everyone in the box can be chosen, even the 17 outside both circles.

2. 2

   $P(F\cap S)=\frac{8}{50}$ — “Both” is the overlap only: 8 students.

3. 3

   $\frac{8}{50}=\frac{4}{25}$ — Divide top and bottom by 2. Either form gets the mark.

4. 4

   $P(F)=\frac{14+8}{50}$ — The French circle has two regions: French only and the overlap.

5. 5

   $\frac{22}{50}=\frac{11}{25}$ — 14 + 8 = 22. Then divide top and bottom by 2.

6. 6

   $P(F\cup S)=\frac{14+8+11}{50}$ — “French or Spanish” means in at least one circle: three regions, with the overlap counted once.

7. 7

   $=\frac{33}{50}$ — 14 + 8 + 11 = 33. The 17 outside both circles are left out.

8. 8

   $P(F\cap S^{\prime })=\frac{14}{50}=\frac{7}{25}$ — “French but not Spanish” is French only. Drop the overlap, because those 8 also study Spanish.

Worked example

1. 1

   $\text{total}=15$ — Count every number in the box, outside the circles too. 15 goes on the bottom.

2. 2

   $A\cap B=\{3,9,15\}$ — The overlap holds the odd multiples of 3. List them, then count: 3 numbers.

3. 3

   $P(A\cap B)=\frac{3}{15}=\frac{1}{5}$ — 3 numbers fit out of 15. Divide top and bottom by 3.

4. 4

   $A\cap B^{\prime }=\{6,12\}$ — “In A but not B” is A only: the multiples of 3 that are even.

5. 5

   $P(A\cap B^{\prime })=\frac{2}{15}$ — 2 numbers fit out of 15.

6. 6

   $(A\cup B{)}^{\prime }=\{2,4,8,10,14\}$ — Outside both circles: not a multiple of 3 and not odd.

7. 7

   $P((A\cup B{)}^{\prime })=\frac{5}{15}=\frac{1}{3}$ — 5 numbers fit out of 15. Divide top and bottom by 5.

Worked example

1. 1

   $P(S^{\prime })=\frac{14+17}{50}$ — “Not Spanish” is every region outside circle S: French only and outside both.

2. 2

   $\frac{31}{50}$ — 14 + 17 = 31. It does not simplify, so leave it.

3. 3

   $1-\frac{19}{50}=\frac{31}{50}$ — Check another way: all of S is 8 + 11 = 19, so take its share away from 1.

4. 4

   $P((F\cup S{)}^{\prime })=\frac{17}{50}$ — “Neither” is outside both circles only: 17 students.

5. 5

   $P(F\cup S^{\prime })=\frac{14+8+17}{50}$ — In F, or not in S: French only, the overlap and outside. Only Spanish only is left out.

6. 6

   $=\frac{39}{50}$ — 14 + 8 + 17 = 39. Check: 50 − 11 = 39.

Worked example

1. 1

   $\text{French: }14+8=22$ — You choose only from the French students, so the French circle is the new total.

2. 2

   $\text{also Spanish: }8$ — Of those 22, the ones who also study Spanish sit in the overlap.

3. 3

   $P=\frac{8}{22}$ — 8 who fit, out of the 22 you choose from.

4. 4

   $=\frac{4}{11}$ — Divide top and bottom by 2.

Worked example

1. 1

   $12+15+9+6+4+8+3+23=80$ — Add all eight regions. Everyone can be chosen, including the 23 in no circle.

2. 2

   $P(\text{exactly two})=\frac{6+4+8}{80}$ — “Exactly two” is the three regions where only two circles overlap. The centre is left out.

3. 3

   $=\frac{18}{80}=\frac{9}{40}$ — 6 + 4 + 8 = 18. Then divide top and bottom by 2.

4. 4

   $n(G)=15+6+8+3=32$ — Now a golfer is chosen. Add the four regions inside circle G: the new total.

5. 5

   $\text{also squash: }6+3=9$ — Inside G, the people who also play squash are “S and G only” and the centre.

6. 6

   $P=\frac{9}{32}$ — 9 who fit, out of the 32 golfers you choose from.