Relative Frequency and Expectation

Worked example

1. 1

   $80\text{ spins, }22\text{ stars}$ — The results. The spinner has 4 equal parts and 1 is a star. Each spin is one trial, so there are 80 trials.

2. 2

   $\frac{22}{80}$ — Put the times it landed on a star over the number of trials.

3. 3

   $\frac{22}{80}=0.275$ — Divide: 22 ÷ 80 = 0.275. This is the relative frequency of a star.

4. 4

   $\text{fair: }\frac{1}{4}=0.25$ — A fair spinner lands on the star 1 time in 4. Work it out so you have something to compare with.

5. 5

   $0.275\approx 0.25$ — The two are close, so after 80 spins the spinner looks fair.

Worked example

1. 1

   $9+14+11+16=50$ — Add all the frequencies. Every spin is counted once, so this is the number of trials.

2. 2

   $\frac{16}{50}$ — A score of 4 came 16 times. Put that over the trials.

3. 3

   $=0.32$ — 16 ÷ 50 = 0.32. This is the relative frequency of scoring 4.

Worked example

1. 1

   $P(\text{win})=0.35$ — Start with what you know: the chance that Sami wins one game.

2. 2

   $0.35\times 160$ — Each game gives the same chance of a win, so for 160 games multiply by 160.

3. 3

   $=56$ — 0.35 × 160 = 56. Sami should win about 56 games.

4. 4

   $P(\text{lose})=1-0.35=0.65$ — Each game is a win or a loss, so the two chances add to 1. Take 0.35 away from 1.

5. 5

   $0.65\times 160=104$ — Multiply P(lose) by the 160 games: about 104 losses.

6. 6

   $56+104=160$ — Check: the wins and the losses together must make all 160 games.

Worked example

1. 1

   $\frac{3}{8}\times 40$ — Same rule: probability times trials. This time the probability is a fraction.

2. 2

   $40÷8=5$ — Divide by the bottom number first: one eighth of 40 games is 5 games.

3. 3

   $5\times 3=15$ — Then multiply by the top number: three eighths of 40 is about 15 wins.

Worked example

1. 1

   $0.3+0.18=0.48$ — Add the probabilities you know. All four probabilities must add up to 1.

2. 2

   $1-0.48=0.52$ — Take that from 1. What is left, 0.52, is shared by 2 and 4.

3. 3

   $x=0.52÷2=0.26$ — 2 and 4 are equally likely, so each one gets half of 0.52.

4. 4

   $0.26\times 150=39$ — Now probability times trials: expect about 39 of the 150 spins to land on 4.

Worked example

1. 1

   $1-0.16=0.84$ — Lose and draw share what is left after winning, because the three add to 1.

2. 2

   $\text{draw}=x,\text{lose}=2x$ — Losing is twice as likely as drawing. Call the draw x, so the lose is 2x.

3. 3

   $x+2x=3x=0.84$ — Together they make the 0.84 that is left. That is 3 equal parts.

4. 4

   $x=0.84÷3=0.28$ — Divide by 3 to find one part. This is P(draw).

5. 5

   $2\times 0.28=0.56$ — Losing is two parts, so P(lose) = 0.56.

6. 6

   $0.16+0.56+0.28=1$ — Check: the three probabilities add to 1.

Worked example

1. 1

   $P(6)=\frac{1}{6}$ — A fair dice has 6 equally likely faces, so the chance of a six is one in six.

2. 2

   $\frac{1}{6}\times 120=20$ — Probability times trials: a fair dice gives about 20 sixes in 120 rolls.

3. 3

   $\frac{41}{120}\approx 0.34$ — The relative frequency of a six in the results. A fair dice gives about 0.17.

4. 4

   $41\text{ is far more than }20$ — About double what a fair dice gives, so the dice could be biased.

Worked example

1. 1

   $400÷5=80$ — A fair spinner shares the 400 spins equally, so each score should come up about 80 times.

2. 2

   $78,83,79,81,79\approx 80$ — Every count is close to 80, so the spinner looks fair. Say so with the numbers.

Worked example

1. 1

   $\frac{9}{40}$ — 9 of the 40 students asked cycle. This is the relative frequency of cycling.

2. 2

   $\frac{9}{40}\times 1200$ — Assume the whole school is like the sample, so multiply by the 1200 students.

3. 3

   $1200÷40=30$ — Divide first to keep the numbers small: the school is 30 times as big as the sample.

4. 4

   $9\times 30=270$ — So there are about 30 times as many cyclists: about 270 students.