FP1–FP3, FP5, HP1–HP3, HP5 - Experiments and relative frequency

FP1–FP3, FP5, HP1–HP3, HP5 - Experiments and relative frequency

Record observed outcomes, use their proportions to estimate probabilities, and predict future counts without treating predictions as guarantees. FP1 and HP1 cover frequency tables and trees; FP2–FP3, FP5 and their Higher counterparts connect experimental results, theoretical chances and expected frequencies.

Recording experiments

If a fair coin is tossed 100100 times, must exactly 5050 results be heads? To compare what happened with what a model predicts, first record the actual outcomes accurately.

Frequency

The frequency of an outcome is the number of times it was observed. A frequency table or frequency tree records counts, not probabilities.

If a trial consists of two coin tosses, keep their order. HTHT means first heads, then tails; THTH is a different outcome. The complete possibilities are HH,HT,TH,TTHH,HT,TH,TT, whether or not the coin is fair.

Worked example 1: completing a frequency table

Question. In 5050 two-toss trials, the observed frequencies are:

OutcomeHHHHHTHTTHTHTTTT
Frequency161612121010xx

Find xx and the number of trials starting with heads.

Plan. Use the overall total for the missing outcome, then combine the two outcomes starting with heads.

x=50−16−12−10=12first heads=16+12=28\begin{aligned} x&=50-16-12-10=12\\ \text{first heads}&=16+12=28 \end{aligned}

Answer. TTTT occurred 1212 times; 2828 trials started with heads.

Check. 16+12+10+12=5016+12+10+12=50. The other 10+12=2210+12=22 trials started with tails, and 28+22=5028+22=50.

The same observations can be organised by stages instead of complete outcome labels.

Worked example 2: completing a frequency tree

Question. Of 5050 trials, 2828 start with heads and 2222 start with tails. Within the first group, 1616 finish with heads. Within the second group, 1010 finish with heads. Find both groups' counts that finish with tails.

Plan. At each split, subtract from that branch's own parent count.

HT=28−16=12TT=22−10=12\begin{aligned} HT&=28-16=12\\ TT&=22-10=12 \end{aligned}

Answer. HT=12HT=12, TT=12TT=12.

Check. The children sum to each parent: 16+12=2816+12=28, 10+12=2210+12=22. The parents sum to the root 5050.

[DIAGRAM: asset_name: Observed two-toss frequencies; asset_slug: observed-frequency-tree; caption: Child counts add to their own parent at each split.; description: A frequency tree starts at fifty trials, splits into first heads twenty-eight and first tails twenty-two, then into HH sixteen, HT twelve, TH ten and TT twelve.]
Child counts add to their own parent at each split.

To find all trials ending in heads, combine HHHH and THTH: 16+10=2616+10=26. Read the stage names carefully: “first heads” and “second heads” describe different groups.

Watch out

The missing child of the 2828-trial branch is 28−1628-16, not 50−1650-16. Use the total for the group being split.

Complete a fresh set of observed counts.

Relative frequency

An experiment can estimate a probability even when the device's theoretical chances are unknown. Use the proportion of trials producing the event.

Relative frequency

The relative frequency of an event is

observed event frequencytotal number of trials\frac{\text{observed event frequency}}{\text{total number of trials}}

It is an experimental proportion and can be used as an estimate of the event's probability.

Use all relevant trials conducted under the same conditions. Do not select a favourable run or assume that the number of colour labels makes a spinner fair.

Worked example 3: an experimental estimate

Question. A spinner lands on red 2828 times in 8080 spins. Estimate the probability of red.

Plan. Divide red frequency by the total number of spins.

P(red)≈2880=0.35P(\text{red})\approx\frac{28}{80}=0.35

Answer. An estimate is 0.350.35.

Check. 0.35×80=280.35\times80=28, recovering the observed count. The estimate lies between 00 and 11.

Extend the experiment with a different-sized batch. Combine counts before dividing.

Worked example 4: combining experiments

Question. The same spinner is spun another 120120 times under the same conditions and lands on red 5454 times. Combine this with the first 2828 reds in 8080 spins to improve the estimate.

Plan. Add event counts and trial counts separately.

red frequency=28+54=82total trials=80+120=200P(red)≈82200=0.41\begin{aligned} \text{red frequency}&=28+54=82\\ \text{total trials}&=80+120=200\\ P(\text{red})&\approx\frac{82}{200}=0.41 \end{aligned}

Answer. 0.410.41.

Check. The batch proportions are 0.350.35 and 0.450.45, so the combined estimate lies between them and nearer the larger batch's 0.450.45.

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