Combined Events with Two-Way Tables

Worked example

1. 1

   $\text{girls}=80-38=42$ — Everyone is a boy or a girl, so take the 38 boys away from the 80 students.

2. 2

   $\text{boys, car}=16-7=9$ — The car column has only one gap. 16 come by car and 7 of them are girls, so the other 9 are boys.

3. 3

   $\text{boys, bus}=38-12-9=17$ — Now the boys row has only one gap. The boys must add to 38, so take away the 12 who walk and the 9 who come by car.

4. 4

   $\text{girls, walk}=27-12=15$ — The walk column has one gap. 27 walk and 12 of them are boys, so 15 are girls.

5. 5

   $\text{bus total}=17+20=37$ — Add down the bus column: 17 boys and 20 girls.

6. 6

   $15+20+7=42$ — Check: the girls row must add to the 42 girls from the first line. It does, so the table is right.

Worked example

1. 1

   $P(\text{girl, walks})=\frac{15}{80}$ — A girl who walks is one cell: 15. Any of the 80 students could be picked, so divide by 80.

2. 2

   $=\frac{3}{16}$ — Simplify: divide the top and the bottom by 5.

3. 3

   $P(\text{bus})=\frac{37}{80}$ — Anyone who takes the bus counts, so use the bus total, 37, over all 80. It does not simplify.

4. 4

   $P(\text{not car})=\frac{80-16}{80}$ — Not car means everyone except the 16 who come by car.

5. 5

   $=\frac{64}{80}=\frac{4}{5}$ — 80 − 16 = 64. Then divide the top and the bottom by 16.

6. 6

   $\text{a girl: }P(\text{car})=\frac{7}{42}$ — The question picks one of the girls, so only the 42 girls can be picked. 7 of them come by car.

7. 7

   $=\frac{1}{6}$ — Simplify: divide the top and the bottom by 7.

Worked example

1. 1

   $B=1:3,5,7$ — The row for $B=1$: add 1 to each number on A, so 2 + 1, 4 + 1 and 6 + 1.

2. 2

   $B=2:4,6,8$ — The next row: add 2 to each number on A.

3. 3

   $B=3:5,7,9$ — Add 3 to each number on A.

4. 4

   $B=4:6,8,10$ — Add 4 to each number on A. Every row is done, so the table is full.

5. 5

   $3\times 4=12\text{ cells}$ — A has 3 sections and B has 4, so there are 12 cells. Each cell is equally likely.

6. 6

   $P(7)=\frac{2}{12}$ — Count the cells that show 7: one in the $B=1$ row and one in the $B=3$ row.

7. 7

   $=\frac{1}{6}$ — Simplify: divide the top and the bottom by 2.

Worked example

1. 1

   $A=1:0,2,3,5$ — The row for $A=1$. Take the smaller number from the bigger one each time: 1 − 1, 3 − 1, 4 − 1, 6 − 1.

2. 2

   $A=2:1,1,2,4$ — The row for $A=2$: 2 − 1, 3 − 2, 4 − 2, 6 − 2. It is 3 − 2, not 2 − 3, because a difference is never negative.

3. 3

   $A=2:1,1,2,4$ — Spinner A shows 2 on two sections, so write the 2 row twice. Each section is its own equally likely outcome.

4. 4

   $3\times 4=12\text{ cells}$ — 3 sections on A times 4 sections on B. Every cell is equally likely.

5. 5

   $P(1)=\frac{4}{12}$ — Count the cells that show 1: two in each 2 row, so 4.

6. 6

   $=\frac{1}{3}$ — Simplify: divide the top and the bottom by 4.

Worked example

1. 1

   $\text{TA, TB, TC}$ — Fix the first drink, tea, and pair it with every snack in turn.

2. 2

   $\text{JA, JB, JC}$ — Move to the next drink, juice, and run through the same three snacks.

3. 3

   $\text{WA, WB, WC}$ — The last drink, water, with the same three snacks.

4. 4

   $3\times 3=9$ — Check the count: 3 drinks times 3 snacks makes 9 pairs, and the list has 9.

Worked example

1. 1

   $P(\text{bus})\times P(\text{train})=0.56$ — Both on time means multiply, and the question says the answer is 0.56.

2. 2

   $0.8\times P(\text{train})=0.56$ — Put in the bus probability, 0.8.

3. 3

   $P(\text{train})=0.56÷0.8=0.7$ — Divide both sides by 0.8 to undo the multiply.

4. 4

   $P(\text{bus late})=1-0.8=0.2$ — Late is the opposite of on time, and the two must add to 1.

5. 5

   $P(\text{train late})=1-0.7=0.3$ — The same idea for the train.

6. 6

   $P(\text{both late})=0.2\times 0.3=0.06$ — Both late is two events together, so multiply.

7. 7

   $P(\text{at least one on time})=1-0.06$ — Every outcome has at least one on time except both late. So take that one case away from 1.

8. 8

   $=0.94$ — 1 − 0.06 = 0.94.