In order to predict the future, we sometimes need to determine the probability by running experiments, or looking at data that has already been collected. This is called experimental probability, since we determine the probability of each outcome by looking at past events. It can also be called relative frequency.
Imagine we have a "loaded" die, where a weight is placed inside the die opposite the face that the cheater wants to come up the most (in this case, the $6$6):
If the die is made like this, the probability of each outcome is no longer equal, and we cannot say that the probability of rolling any particular face is $\frac{1}{6}$16.
Instead we will need to roll the die many times and record our results, and use these results to predict the future. Here are the results of an experiment where the die was rolled $200$200 times:
Result | Number of rolls |
---|---|
$1$1 | $11$11 |
$2$2 | $19$19 |
$3$3 | $18$18 |
$4$4 | $18$18 |
$5$5 | $20$20 |
$6$6 | $114$114 |
We can now try to predict the future using this experimental data, and the following formula:
$\text{Experimental probability of event}=\frac{\text{Number of times event occurred in experiments}}{\text{Total number of experiments}}$Experimental probability of event=Number of times event occurred in experimentsTotal number of experiments
Here is the table again, with the experimental probability of each face listed as a percentage:
Result | Number of rolls | Experimental probability |
---|---|---|
$1$1 | $11$11 | $5.5%$5.5% |
$2$2 | $19$19 | $9.5%$9.5% |
$3$3 | $18$18 | $9%$9% |
$4$4 | $18$18 | $9%$9% |
$5$5 | $20$20 | $10%$10% |
$6$6 | $114$114 | $57%$57% |
A normal die has around $17%$17% chance of rolling a $6$6, but this die rolls a $6$6 more than half the time!
Sometimes our "experiments" involve looking at historical data instead. For example, we can't run hundreds of Eurovision Song Contests to test out who would win, so instead we look at past performance when trying to predict the future.
The following table shows the winner of the Eurovision Song Contest from 1999 to 2018:
Year | Winning country | Year | Winning country |
---|---|---|---|
1999 | Sweden | 2009 | Norway |
2000 | Denmark | 2010 | Germany |
2001 | Estonia | 2011 | Azerbaijan |
2002 | Latvia | 2012 | Sweden |
2003 | Turkey | 2013 | Denmark |
2004 | Ukraine | 2014 | Austria |
2005 | Greece | 2015 | Sweden |
2006 | Finland | 2016 | Ukraine |
2007 | Serbia | 2017 | Portugal |
2008 | Russia | 2018 | Israel |
(a) What is the experimental probability that Sweden will win the next Eurovision Song Contest?
We think of each contest as an "experiment", and there are $20$20 in total. The winning country is the event, and we can tell that $3$3 of the contests were won by Sweden. So using the same formula as above,
$\text{Experimental probability of event}=\frac{\text{Number of times event occurred in experiments}}{\text{Total number of experiments}}$Experimental probability of event=Number of times event occurred in experimentsTotal number of experiments
the experimental probability is $\frac{3}{20}$320, which is $15%$15%.
(b) How many of the next $50$50 contests can Sweden expect to win?
We can calculate this by multiplying the experimental probability of an event by the number of trials. In this case Sweden can expect to win
$\frac{3}{20}\times50=\frac{150}{20}$320×50=15020 contests
This rounds to $8$8 contests out of the next $50$50.
$500$500 cables were tested at a factory, and $76$76 were found to be faulty.
(a) What is the experimental probability that a cable at this factory will be faulty?
$P\left(\text{Faulty}\right)$P(Faulty) | $=$= | $\frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$Number of favourable outcomesTotal number of trials |
$=$= | $\frac{76}{500}$76500 | |
$=$= | $0.152$0.152 |
(b) If $1500$1500 more cables were tested, how many would you expect to be faulty?
Now that we know that approximately $15.2%$15.2% are faulty (from our experimental data), we could expect the same percentage to be faulty from any amount.
So:
$15.2%$15.2% of $1500$1500$=$=$0.152\times1500=228$0.152×1500=228
We could expect $228$228 to be faulty from $1500$1500 cables.
A die is rolled $60$60 times and the results are recorded in the following table:
Number | Frequency |
---|---|
$1$1 | $8$8 |
$2$2 | $10$10 |
$3$3 | $8$8 |
$4$4 | $10$10 |
$5$5 | $10$10 |
$6$6 | $14$14 |
What is the relative frequency of rolling a $6$6 with this die? Express your answer in simplest form.
What is the relative frequency of rolling a $3$3 or higher with this die? Express your answer in simplest form.
What is the relative frequency of rolling a $3$3 or lower with this die? Express your answer in simplified form.
A retail store served $773$773 customers in October, and there were $44$44 complaints during that month.
Determine, as a percentage, the experimental probability that a customer submits a complaint.
Round your answer to the nearest whole percent.
An insurance company found that in the past year, of the $2558$2558 claims made, $1493$1493 of them were from drivers under the age of 25.
Give your answers to the following questions as percentages, rounded to the nearest whole percent.
What is the experimental probability that a claim is filed by someone under the age of 25?
What is the experimental probability that a claim is filed by someone 25 or older?
The experimental probability that a commuter uses public transport is $50%$50%.
Out of $500$500 commuters, how many would you expect to use public transport?
This frequency graph shows the number of people that were served at a furniture store, and the length of time it took to serve them.
What is the probability that someone was served in under $40$40 minutes?
What is the probability that someone had to wait at least $50$50 minutes to be served?