Experimental and theoretical probabilities
🎯 In this topic you will
- Use experiments to calculate experimental probabilities
- Calculate experimental probabilities and compare them to theoretical probabilities
🧠 Key Words
- experiment
- experimental probability
- relative frequency
- theoretical probability
- trial
Show Definitions
- experiment: A practical activity or test carried out to investigate outcomes or gather data.
- experimental probability: The probability of an event based on the results observed from experiments or trials.
- relative frequency: The ratio of the number of times an event occurs to the total number of trials, used to estimate probability.
- theoretical probability: The probability of an event calculated based on mathematical reasoning or known possible outcomes, without experiments.
- trial: A single attempt or repetition of an experiment.
When outcomes are not equally likely, you can use an experiment to estimate probabilities.
If you drop a drawing pin, it can land point up or point down.
You cannot assume that these two outcomes are equally likely.
What is the probability of the drawing pin landing point up?
You can do an experiment to find the experimental probability of the drawing pin landing point up.
To do this, drop a large number of drawing pins.
Work out the relative frequency of point up = $\dfrac{\text{frequency of point up}}{\text{total number of drawing pins}}$
This tells you an experimental probability for point up.
You can also get the answer to part b of Worked example 13.3 by finding $100\% - 39\% = 61\%$.
When you do an experiment to find a probability, the answer might not be exactly the same each time. A large number of trials gives a more reliable answer than a small number of trials.
❓ EXERCISES
1. A survey of $400$ cars travelling along a road is carried out.
The results show that $140$ cars were travelling faster than $60 \,\text{km/h}$.
Find the experimental probability that the next car will be:
a. travelling faster than $60 \,\text{km/h}$
b. travelling at $60 \,\text{km/h}$ or less
👀 Show answer
a. Experimental probability of travelling faster than $60 \,\text{km/h}$:
$\dfrac{140}{400} = 0.35 = 35\%$
b. Experimental probability of travelling at $60 \,\text{km/h}$ or less:
$\dfrac{260}{400} = 0.65 = 65\%$
❓ EXERCISES
2. There are $320$ students in a school. $16$ students travel to school by car. $96$ students walk to school.
Find the experimental probability that a particular student:
a. travels by car to school
b. walks to school
c. does not walk to school
d. does not walk or travel by car to school
👀 Show answer
a.$\dfrac{16}{320}=\dfrac{1}{20}=0.05=5\%$
b.$\dfrac{96}{320}=\dfrac{3}{10}=0.30=30\%$
c. Not walking: $320-96=224$ → $\dfrac{224}{320}=\dfrac{7}{10}=0.70=70\%$
d. Neither walk nor car: $320-(96+16)=208$ → $\dfrac{208}{320}=\dfrac{13}{20}=0.65=65\%$
3. Mrs Patel drives to work each day. Sometimes she must stop at a set of traffic lights. In the past $50$ working days she has stopped at this set of traffic lights $32$ times.
a. Find the experimental probability that Mrs Patel will have to stop at this set of lights tomorrow.
b. Find the experimental probability that she will not have to stop at the lights next Wednesday.
👀 Show answer
a.$\dfrac{32}{50}=0.64=64\%$
b.$\dfrac{50-32}{50}=\dfrac{18}{50}=0.36=36\%$
4. Jasmine goes to school five days a week. In the past eight weeks she has been late for school on three days. Estimate the probability that tomorrow Jasmine will be:
a. late for school b. on time
👀 Show answer
Total days: $8\times5=40$.
a. Late: $\dfrac{3}{40}=0.075=7.5\%$
b. On time: $\dfrac{37}{40}=0.925=92.5\%$
5. Carlos looks at the weather records for November for his town. Over the past five years ($150$ days) there has been rain on $36$ days in November. Use this information to estimate the following probabilities. Write your answers as percentages.
a. Estimate the probability that it will rain on 1 November next year.
b. Estimate the probability that it will not rain on 30 November next year.
👀 Show answer
a. Rain: $\dfrac{36}{150}=0.24=24\%$
b. Not rain: $\dfrac{150-36}{150}=\dfrac{114}{150}=0.76=76\%$
6. Arun says:
a. How did Arun work out this probability?
b. This is not a good way to estimate the probability. Why?
c. Compare your answer to part b with a partner’s answer. Have you both given the same explanation? Can you improve your explanation?
👀 Show answer
a. He used the relative frequency of wins: $\dfrac{15}{20}=0.75=75\%$.
b. A single future match may not be similar to past matches (different opponent, home/away, injuries, etc.), and the sample size is small; the estimate may be unreliable.
c. Answers will vary. Discuss and refine reasons such as small sample size and changing conditions.
❓ EXERCISES
7. Here are the results of a survey of $240$ students attending a college.
| Item | Has a mobile phone | Has a computer in their bedroom | Wants to be in a band | Is a member of a sports team |
|---|---|---|---|---|
| Number of students | $232$ | $167$ | $92$ | $68$ |
a. Find the experimental probability that a student chosen at random from the college:
i. has a mobile phone ii. has a computer in their bedroom
iii. is not a member of a sports team
Give each answer as a percentage, rounded to the nearest whole number.
b. Why is the following argument incorrect?
A good estimate of the probability that a student at the college wants to be in a band or is a member of a sports team is $\dfrac{92+68}{240}=\dfrac{160}{240}=\dfrac{2}{3}$ or $67\%$.
👀 Show answer
a. i.$\dfrac{232}{240}=0.966\overline{6}\approx97\%$
a. ii.$\dfrac{167}{240}\approx0.6958\approx70\%$
a. iii. Not a member: $\dfrac{240-68}{240}=\dfrac{172}{240}\approx0.7167\approx72\%$
b. The sets “wants to be in a band” and “is a member of a sports team” may overlap. Adding $92$ and $68$ double-counts students in both groups; without the intersection you cannot use simple addition (use inclusion–exclusion).
8. Varun flips a coin. The two possible outcomes are heads and tails.
a. If the outcomes are equally likely, what are the theoretical probabilities of each outcome?
b. Varun’s results are shown in the table.
| Outcome | heads | tails | total |
|---|---|---|---|
| Frequency | $24$ | $16$ | $40$ |
Use the results to find the relative frequency of each outcome.
c. Varun’s friend Toby says that Varun is not flipping the coin fairly because the probabilities from the experiment are incorrect. What does Toby mean? Do you think Toby is correct?
👀 Show answer
a. Heads: $\dfrac{1}{2}=0.5=50\%$, Tails: $\dfrac{1}{2}=0.5=50\%$
b. Heads: $\dfrac{24}{40}=0.6=60\%$, Tails: $\dfrac{16}{40}=0.4=40\%$
c. Toby compares the experimental results ($0.6$, $0.4$) with the theoretical probabilities ($0.5$, $0.5$) and claims the coin isn’t fair. But experimental results can differ from theory for a small number of trials; with more flips the relative frequencies should approach $50\%$ each (law of large numbers).
9. A bag contains one white ball, one black ball and some red balls. A ball is chosen without looking.
a. If there are three red balls, work out the theoretical probability of choosing each colour.
Daniella takes out one ball, records the colour and replaces it in the bag. She does this $50$ times. She records her results in a table.
| Outcome | white | black | red | total |
|---|---|---|---|---|
| Frequency | $6$ | $8$ | $36$ | $50$ |
b. Use the results to find the experimental probability of choosing each of the three colours.
c. Daniella knows that there are an odd number of red balls. What is the most likely number? Give a reason for your answer.
👀 Show answer
a. With three red balls the bag has $5$ balls in total. White: $\dfrac{1}{5}=0.2=20\%$, Black: $\dfrac{1}{5}=0.2=20\%$, Red: $\dfrac{3}{5}=0.6=60\%$.
b. From the table: White: $\dfrac{6}{50}=0.12=12\%$; Black: $\dfrac{8}{50}=0.16=16\%$; Red: $\dfrac{36}{50}=0.72=72\%$.
c. Let the number of red balls be $r$. Then the red probability is $\dfrac{r}{r+2}$. We want an odd$r$ with $\dfrac{r}{r+2}\approx0.72$. For $r=5$, $\dfrac{5}{7}\approx0.714$; for $r=7$, $\dfrac{7}{9}\approx0.778$. The closest is $r=5$, so the most likely number of red balls is $5$ (because it gives a probability closest to the experimental value).
❓ EXERCISES
7. Here are the results of a survey of $240$ students attending a college.
| Item | Has a mobile phone | Has a computer in their bedroom | Wants to be in a band | Is a member of a sports team |
|---|---|---|---|---|
| Number of students | $232$ | $167$ | $92$ | $68$ |
a. Find the experimental probability that a student chosen at random from the college:
i. has a mobile phone ii. has a computer in their bedroom
iii. is not a member of a sports team
Give each answer as a percentage, rounded to the nearest whole number.
b. Why is the following argument incorrect?
A good estimate of the probability that a student at the college wants to be in a band or is a member of a sports team is $\dfrac{92+68}{240}=\dfrac{160}{240}=\dfrac{2}{3}$ or $67\%$.
👀 Show answer
a. i.$\dfrac{232}{240}=0.966\overline{6}\approx97\%$
a. ii.$\dfrac{167}{240}\approx0.6958\approx70\%$
a. iii. Not a member: $\dfrac{240-68}{240}=\dfrac{172}{240}\approx0.7167\approx72\%$
b. The sets “wants to be in a band” and “is a member of a sports team” may overlap. Adding $92$ and $68$ double-counts students in both groups; without the intersection you cannot use simple addition (use inclusion–exclusion).
8. Varun flips a coin. The two possible outcomes are heads and tails.
a. If the outcomes are equally likely, what are the theoretical probabilities of each outcome?
b. Varun’s results are shown in the table.
| Outcome | heads | tails | total |
|---|---|---|---|
| Frequency | $24$ | $16$ | $40$ |
Use the results to find the relative frequency of each outcome.
c. Varun’s friend Toby says that Varun is not flipping the coin fairly because the probabilities from the experiment are incorrect. What does Toby mean? Do you think Toby is correct?
👀 Show answer
a. Heads: $\dfrac{1}{2}=0.5=50\%$, Tails: $\dfrac{1}{2}=0.5=50\%$
b. Heads: $\dfrac{24}{40}=0.6=60\%$, Tails: $\dfrac{16}{40}=0.4=40\%$
c. Toby compares the experimental results ($0.6$, $0.4$) with the theoretical probabilities ($0.5$, $0.5$) and claims the coin isn’t fair. But experimental results can differ from theory for a small number of trials; with more flips the relative frequencies should approach $50\%$ each (law of large numbers).
9. A bag contains one white ball, one black ball and some red balls. A ball is chosen without looking.
a. If there are three red balls, work out the theoretical probability of choosing each colour.
Daniella takes out one ball, records the colour and replaces it in the bag. She does this $50$ times. She records her results in a table.
| Outcome | white | black | red | total |
|---|---|---|---|---|
| Frequency | $6$ | $8$ | $36$ | $50$ |
b. Use the results to find the experimental probability of choosing each of the three colours.
c. Daniella knows that there are an odd number of red balls. What is the most likely number? Give a reason for your answer.
👀 Show answer
a. With three red balls the bag has $5$ balls in total. White: $\dfrac{1}{5}=0.2=20\%$, Black: $\dfrac{1}{5}=0.2=20\%$, Red: $\dfrac{3}{5}=0.6=60\%$.
b. From the table: White: $\dfrac{6}{50}=0.12=12\%$; Black: $\dfrac{8}{50}=0.16=16\%$; Red: $\dfrac{36}{50}=0.72=72\%$.
c. Let the number of red balls be $r$. Then the red probability is $\dfrac{r}{r+2}$. We want an odd$r$ with $\dfrac{r}{r+2}\approx0.72$. For $r=5$, $\dfrac{5}{7}\approx0.714$; for $r=7$, $\dfrac{7}{9}\approx0.778$. The closest is $r=5$, so the most likely number of red balls is $5$ (because it gives a probability closest to the experimental value).
❓ EXERCISES
10. Work with a partner to answer this question.
This table shows $200$ random digits generated by a spreadsheet.
| $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ | $15$ | $16$ | $17$ | $18$ | $19$ | $20$ |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| $0$ | $2$ | $7$ | $5$ | $0$ | $8$ | $6$ | $9$ | $0$ | $7$ | $8$ | $0$ | $3$ | $2$ | $9$ | $2$ | $9$ | $7$ | $4$ | $7$ |
| $5$ | $7$ | $3$ | $0$ | $2$ | $9$ | $8$ | $4$ | $5$ | $3$ | $9$ | $6$ | $4$ | $9$ | $7$ | $5$ | $3$ | $1$ | $1$ | $5$ |
| $2$ | $8$ | $3$ | $0$ | $4$ | $3$ | $7$ | $6$ | $5$ | $5$ | $0$ | $5$ | $7$ | $9$ | $4$ | $1$ | $8$ | $8$ | $1$ | $4$ |
| $8$ | $9$ | $2$ | $6$ | $5$ | $4$ | $1$ | $0$ | $9$ | $6$ | $3$ | $4$ | $1$ | $0$ | $1$ | $5$ | $4$ | $9$ | $2$ | $3$ |
| $2$ | $7$ | $9$ | $7$ | $1$ | $0$ | $1$ | $5$ | $7$ | $1$ | $6$ | $6$ | $6$ | $2$ | $2$ | $6$ | $2$ | $6$ | $3$ | $4$ |
| $8$ | $1$ | $8$ | $0$ | $1$ | $4$ | $2$ | $0$ | $3$ | $7$ | $6$ | $0$ | $1$ | $9$ | $0$ | $9$ | $6$ | $7$ | $8$ | $2$ |
| $0$ | $5$ | $9$ | $0$ | $6$ | $1$ | $8$ | $7$ | $9$ | $4$ | $8$ | $0$ | $2$ | $7$ | $9$ | $3$ | $8$ | $3$ | $9$ | $5$ |
| $2$ | $4$ | $6$ | $1$ | $7$ | $3$ | $8$ | $0$ | $1$ | $1$ | $4$ | $1$ | $0$ | $7$ | $3$ | $4$ | $9$ | $1$ | $5$ | $5$ |
| $0$ | $8$ | $9$ | $9$ | $0$ | $6$ | $7$ | $8$ | $2$ | $6$ | $2$ | $6$ | $2$ | $1$ | $2$ | $2$ | $4$ | $9$ | $3$ | $8$ |
| $1$ | $2$ | $4$ | $6$ | $6$ | $3$ | $0$ | $1$ | $1$ | $7$ | $0$ | $8$ | $5$ | $9$ | $0$ | $8$ | $1$ | $5$ | $0$ | $1$ |
a. If every digit from $0$ to $9$ is equally likely, what is the theoretical probability that a digit is $0$?
b. Use the first row of the table as a sample of $20$ digits. Use it to work out the experimental probability that a digit is $0$. Write the answer as a decimal.
c. Now use the first two rows as a sample of $40$ digits. Use it to work out the experimental probability that a digit is $0$.
d. Continue in this way to get samples of $60$, $80$ and so on, up to $200$. In each case, work out the experimental probability that the digit is $0$.
e. Look at the ten experimental probabilities you have calculated so far. What do you notice about them?
f. Choose a different digit and do the same calculations. Comment on your results.
👀 Show answer
a.$\dfrac{1}{10}=0.1$
b. Zeros in row $1$: $4$ → $\dfrac{4}{20}=0.20$
c. Zeros in rows $1$–$2$: $5$ → $\dfrac{5}{40}=0.125$
d. Cumulative counts of zeros and probabilities:
$60$: $\dfrac{7}{60}\approx0.1167$ $80$: $\dfrac{9}{80}=0.1125$ $100$: $\dfrac{10}{100}=0.10$
$120$: $\dfrac{14}{120}\approx0.1167$ $140$: $\dfrac{17}{140}\approx0.1214$ $160$: $\dfrac{19}{160}=0.11875$
$180$: $\dfrac{21}{180}=0.1167$ $200$: $\dfrac{25}{200}=0.125$
e. They fluctuate around $0.1$ and tend to get closer to $0.1$ as the sample size increases (law of large numbers).
f. Answers will vary; for any other digit, the experimental probabilities should also vary around $0.1$ and trend toward $0.1$ as the sample grows.
🧠 Think like a Mathematician
Task: Work on your own to investigate patterns in the digits of $\pi$.
The number $\pi$ is very important in mathematics. When written as a decimal it does not terminate. Here are the first 250 digits of $\pi$:
Investigation:
Is every digit from $0$ to $9$ equally likely, or are some digits more likely than others? Decide how you will investigate this question. Give evidence for any conclusion you reach.
Suggested Method:
- Count how many times each digit (0–9) appears in the first 250 digits of $\pi$.
- Record the frequencies in a table.
- Convert the frequencies into relative frequencies (experimental probabilities).
- Compare these with the theoretical probability of $0.1$ for each digit.
- Decide if the digits appear equally likely or not.
Follow-up Questions:
👀 Show Answer
- 1: The theoretical probability is $\dfrac{1}{10}=0.1$.
- 2: In the first 250 digits, the frequencies of digits are close to each other, but some digits (like 9 or 3) appear slightly more often, while others (like 0 or 1) appear slightly less often.
- 3: The results vary slightly from exact equality, but as the number of digits increases the relative frequencies tend to balance, supporting the idea that digits are equally likely in $\pi$.
You can use equally likely outcomes to calculate probabilities. When this is not possible you can do an experiment.
A spreadsheet is used to simulate throwing a dice $200$ times. Here are the results of the experiment.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | $30$ | $36$ | $37$ | $33$ | $35$ | $29$ |
From the information in the table, we can work out the experimental probabilities:
- The experimental probability of 1 is $\dfrac{30}{200}=0.15$
- The experimental probability of 2 is $\dfrac{36}{200}=0.18$
- The experimental probability of an even number is $\dfrac{36+33+29}{200}=\dfrac{98}{200}=0.49$
We know that each number is equally likely with a fair dice so we can also calculate the theoretical probabilities:
- The theoretical probability of 1 is $\dfrac{1}{6}=0.167$ to 3 d.p.
- The theoretical probability of 2 is $\dfrac{1}{6}=0.167$ to 3 d.p.
- The theoretical probability of an even number is $\dfrac{3}{6}=\dfrac{1}{2}=0.5$
The experimental probabilities and the theoretical probabilities are very similar. This shows that the spreadsheet simulation is reliable.
❓ EXERCISES
1. A learner throws a coin $50$ times. This table shows the results.
| $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ |
|---|---|---|---|---|---|---|---|---|---|
| T | H | T | T | T | H | H | T | H | T |
| H | T | H | H | H | T | H | H | H | T |
| H | H | T | H | H | H | T | H | T | H |
| T | T | T | T | T | H | T | T | T | T |
| T | H | H | T | T | T | H | T | H | T |
a. Use the first row of the table to calculate the experimental probability of a head based on the first $10$ throws.
b. Use the first two rows of the table to calculate the experimental probability of a head based on $20$ throws.
c. In the same way, find the experimental probability of a head based on i. $30$ throws ii. $40$ throws iii. $50$ throws.
d. Compare the experimental probabilities you have found so far with the theoretical probability of a head.
The learner throws the coin another $50$ times. Here are the results.
| $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ |
|---|---|---|---|---|---|---|---|---|---|
| H | H | H | H | T | T | T | H | T | H |
| T | T | T | T | H | T | H | T | T | T |
| T | T | H | T | H | H | T | T | T | T |
| H | T | H | H | H | H | H | H | T | T |
| H | T | T | T | H | H | H | H | T | T |
e. Use the two sets of results to find the experimental probability of a head based on $100$ throws. How close is it to the theoretical probability?
👀 Show answer
a. In row 1: $5$ heads out of $10$ → $P(H)=\tfrac{5}{10}=0.5$
b. In rows 1–2: $12$ heads out of $20$ → $P(H)=\tfrac{12}{20}=0.6$
c. i. First $30$ throws: $17$ heads → $P(H)=\tfrac{17}{30}\approx0.567$
ii. First $40$ throws: $23$ heads → $P(H)=\tfrac{23}{40}=0.575$
iii. All $50$ throws: $28$ heads → $P(H)=\tfrac{28}{50}=0.56$
d. The theoretical probability of a head is $0.5$. The experimental values vary around this but are fairly close.
e. In $100$ throws: total heads = $28+48=76$ → $P(H)=\tfrac{76}{100}=0.76$. This is higher than $0.5$, suggesting the results may not perfectly reflect theory, but with more trials the proportion is expected to move closer to $0.5$.
❓ EXERCISES
2. This spinner has $3$ sectors.
The probability of red, $P(\text{red})=0.6$
The probability of white, $P(\text{white})=0.3$
The probability of blue, $P(\text{blue})=0.1$
Here are the results of $50$ spins.
| Row 1 | Row 2 | Row 3 | Row 4 | Row 5 |
|---|---|---|---|---|
| R W R R R R R B B B | W R W R W R W R R W | R W R W R R R R R R | R W R W R R R B B R | R R R R W R R W R R |
a. Use each row of the table to find an experimental probability of red based on $10$ spins.
b. Find two different sets of $25$ spins and use them to find the experimental probability of red.
c. Use all $50$ spins to find experimental probabilities of red, white and blue.
d. Here are the results of $800$ spins.
| Colour | red | white | blue |
|---|---|---|---|
| Frequency | $489$ | $218$ | $93$ |
Use these results to find experimental probabilities for each colour.
e. Read what Marcus says:
“It is better to use a large number of spins to work out experimental probabilities.”
👀 Show answer
a. Each row has about $6$ reds out of $10$, so $P(\text{red}) \approx 0.6$ for each row.
b. Using two sets of $25$ spins, the proportion of red is close to $0.6$.
c. In $50$ spins, approximate proportions: $P(\text{red}) \approx 0.62$, $P(\text{white}) \approx 0.28$, $P(\text{blue}) \approx 0.10$.
d. In $800$ spins: $P(\text{red}) = \tfrac{489}{800}=0.611$, $P(\text{white}) = \tfrac{218}{800}=0.273$, $P(\text{blue}) = \tfrac{93}{800}=0.116$.
e. Marcus is correct: using a larger sample makes the experimental probabilities closer to the theoretical values.
❓ EXERCISES
3. This question is about throwing six dice together and seeing if there is at least one $6$.
Four learners each threw six dice together a number of times. Here are their results.
| Name | Arun | Sofia | Marcus | Zara |
|---|---|---|---|---|
| Number of throws | $10$ | $20$ | $40$ | $50$ |
| Frequency of at least one $6$ | $7$ | $9$ | $31$ | $36$ |
a. Work out the experimental probability of at least one $6$ for each learner.
b. Combine the four sets of results to get another experimental probability.
c. A computer simulated $500$ throws. There was at least one $6$$333$ times. Work out an experimental probability from this data.
d. In fact, the theoretical probability of throwing at least one $6$ is $0.6651$. Compare the experimental probabilities with the theoretical probability.
👀 Show answer
a. Arun: $\dfrac{7}{10}=0.7$ | Sofia: $\dfrac{9}{20}=0.45$ | Marcus: $\dfrac{31}{40}=0.775$ | Zara: $\dfrac{36}{50}=0.72$.
b. Combined: total throws $10+20+40+50=120$, successes $7+9+31+36=83$ → $\dfrac{83}{120}\approx0.6917$ (about $69.2\%$).
c. Computer: $\dfrac{333}{500}=0.666$ (about $66.6\%$).
d. Theoretical probability (for six dice) is $1-\left(\dfrac{5}{6}\right)^6 \approx 0.6651$. The larger samples (combined and computer) are close to $0.6651$; smaller individual samples vary more (e.g., Sofia low, Marcus high). Increasing trials improves agreement with theory.
❓ EXERCISES
4. Work with one or more other learners on this question.
You learnt about the number $\pi$ in Unit $8$. It is the ratio of the circumference of a circle to its diameter.
The value of $\pi$ is a decimal that does not terminate and has no pattern to its digits.
Here are the first $200$ decimal places of $\pi$:
169 399 375 105 820 974 944 592 307 816 406 286 208 998
628 034 825 342 117 067 982 148 086 513 282 306 647 093
844 609 550 582 231 725 359 408 128 481 117 450 284 102
701 938 521 105 559 644 622 948 954 930 381 96
a. Devise and carry out an experiment to test this statement. Use experimental probabilities and compare them with theoretical probabilities.
b. Describe your experiment and your result. Give a reason for your conclusion.
c. Look at the results of another pair. How do they compare with yours?
👀 Show answer
a. Take the given string of $200$ digits of $\pi$ (or a longer list). Count the frequency of each digit $0,1,\dots,9$. For each digit compute the experimental probability $P(d)=\dfrac{\text{count of }d}{200}$. Compare each with the theoretical probability $0.1$.
b. A valid conclusion explains whether the frequencies are close to $0.1$ and notes small random variation. Reason: sampling variability—over larger samples the relative frequencies tend to move closer to $0.1$ (law of large numbers).
c. Another group’s results should be broadly similar but not identical. Differences are expected from random sampling; bigger samples generally give values nearer to $0.1$.
5. You need a spreadsheet to answer this question. You also need to know how to use it to generate random numbers.
a. Carry out a simulation to model throwing a coin $50$ times. Find the experimental probability of throwing a head and compare it with the theoretical probability.
b. Repeat part a another $5$ times. How much do the experimental probabilities vary?
c. You now have the results of $300$ simulated throws. Use them all to find an experimental probability of throwing a head.
d. Experiment with larger numbers of throws, finding an experimental probability of throwing a head each time. Comment on your results.
👀 Show answer
a. Use the spreadsheet to generate $50$ coin flips and compute $P(H)=\dfrac{\text{heads}}{50}$. Compare with the theoretical probability $0.5$.
b. Repeating gives six estimates of $P(H)$ that typically lie within about $\pm 0.15$ of $0.5$ for samples of size $50$. The values vary from run to run because of randomness.
c. Pool all $300$ throws: $P(H)=\dfrac{\text{total heads}}{300}$. This pooled estimate should be closer to $0.5$ than the separate $50$-throw estimates.
d. As the number of throws increases (e.g., $500$, $1000$, $5000$), the experimental probability should settle nearer to $0.5$ and fluctuate less—consistent with the law of large numbers.