Design and run a conditional experiment
Name a complete trial, a success event and a condition. Match the physical replacement rule, record every trial and then filter the conditioning group
- Define the events Here F means first draw red and L means last draw red. One trial contains every draw, even if the first draw is blue
- Choose a matching device Select individual counters uniformly. For three red and two blue counters, a fair colour coin is wrong: the first red chance is 3/5.
- Separate two resets Replacement governs draws within a trial. Restore the entire bag between complete trials under either rule
- Filter after recording For last red given first red, divide successful eligible trials by all first-red trials, not by all trials
- Interpret uncertainty Repeat runs and compare increasing trial counts. Estimates fluctuate; a larger run does not guarantee a closer estimate every time
Worked example
A simulation of 160 complete two-draw trials records RR = 43, RB = 49, BR = 50 and BB = 18. Estimate last red given first red
- \(\displaystyle n(F)=43+49=92\)Only trials beginning red satisfy the condition
- \(\displaystyle n(L\cap F)=43\)Within that group, RR is a success
- \(\displaystyle \widehat{\Pr}(L\mid F)=43/92\)Use the eligible denominator. This is an estimate, not an exact model probability
- \(\displaystyle \begin{aligned}\widehat{\Pr}(L)&=\frac{43+50}{160}\\&=\frac{93}{160}\end{aligned}\)The unconditional last-red estimate uses every trial. It answers a different question
Common mistakes
Difficulty levels
Questions about this topic
What counts as one trial?+
One complete ordered sequence of two or three draws, with the specified within-trial replacement rule
When do I reset the bag?+
Always before the next complete trial. With replacement, also return each counter before the next draw within the current trial
Does selecting a condition run a new experiment?+
No. It filters the same recorded outcomes. Changing the run number changes the random sequence; increasing the trial count extends the current run
Why can the denominator be smaller than the number of trials?+
Only trials satisfying the condition are eligible. Both their successes and failures belong in the denominator
What if no trial satisfies the condition?+
The conditional estimate is undefined. Collect more data; do not report zero as though the condition had been observed
Does a larger run always get closer?+
No. Larger samples usually reduce fluctuation, but an individual estimate can move away before moving closer
Can I use a fair coin for the colours?+
Only if the required colour probabilities are one-half. For three red and two blue counters, use equally likely individual objects or an equivalent weighted mapping
What should my investigation report include?+
The question, events and condition, model assumptions, replacement rule, device, full trial count, eligible count, successes, estimates across runs, comparison with theory and a limitation
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 10 Mathematics, chapter 4: Probability