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Year 7 Maths Probability

Experimental Probability

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Theory

Experimental probability is found by doing an experiment many times and counting how often an event happens: \(P(E) = \dfrac{\text{frequency of } E}{\text{number of trials}}\). This Year 7 maths page shows how to read frequency tables, compare results with the theoretical probability, predict expected numbers and spot bias.

A trial is one go of an experiment: one toss, one roll or one spin. The frequency of an outcome is the number of times it happened.

The experimental probability (also called the observed probability or relative frequency) is \(P(E) = \dfrac{\text{frequency of } E}{\text{total number of trials}}\). In a table, the total number of trials is the total frequency (add all the frequencies).

The theoretical probability is the chance you expect under fair conditions: a fair coin gives \(P(\text{head}) = \dfrac{1}{2}\), and a fair die gives \(P(\text{any one number}) = \dfrac{1}{6} = 0.1\dot{6}\).

If the division does not end, round it: \(7\) heads in \(30\) tosses gives \(7 \div 30 = 0.2333\ldots \approx 0.23\) (\(2\) decimal places), or about \(23\%\).

The expected number is a prediction: expected \(=\) probability \(\times\) number of trials. A sample’s relative frequency can predict a larger group.

The more trials you do, the closer the relative frequency usually gets to the theoretical probability. A random number generator can run hundreds of trials quickly. If the results stay far from the theory after many trials, the object may be biased (unfair).

Relative frequency of heads as the number of tosses grows A line graph. After 10, 50, 100, 500 and 1000 tosses the relative frequency of heads is 0.3, 0.58, 0.46, 0.518 and 0.493. A dashed line at 0.5 shows the theoretical probability. 0 0.25 0.5 0.75 1 10 50 100 500 1000 number of tosses rel. freq.
A fair coin tossed again and again: \(3\) heads in \(10\), \(29\) in \(50\), \(46\) in \(100\), \(259\) in \(500\), \(493\) in \(1000\). The relative frequency jumps around at first, then settles close to the theoretical \(0.5\) (dashed).

Two rules:

\[ \text{Experimental } P(E) = \dfrac{\text{frequency of } E}{\text{total number of trials}} \]
Experimental P ( E )= frequency of E / total number of trials
\[ \text{Expected} = \text{probability} \times \text{trials} \]
Expected number = probability × number of trials
Experimental and theoretical are rarely equal. \(29\) heads in \(50\) tosses (\(0.58\)) does not prove a coin is unfair. Only a big difference over many trials suggests bias.
Expected means “about”. A fair die rolled \(60\) times is expected to show about \(10\) sixes; \(8\) or \(13\) is normal.

How to find an experimental probability

  1. Add all the frequencies to get the number of trials.
  2. Add the frequencies of the favourable outcomes.
  3. Divide: frequency over trials. Simplify, or round as asked.

How to predict

  1. Choose the probability (theoretical, or experimental from a sample).
  2. Multiply by the number of trials (or the size of the group).
Example 1 — Bottle flips
Tom flipped a water bottle \(60\) times. It landed upright \(15\) times. Find the experimental probability that it lands upright, as a fraction and a decimal.
Solution
\(\dfrac{15}{60}\)\(=\)\(\dfrac{1}{4}\)
\(\)\(=\)\(0.25\)
15/60=1/4=0.25

\(\dfrac{1}{4}\), or \(0.25\)

Example 2 — Reading a frequency table
A die was rolled and the results recorded.
Outcome: \(1, 2, 3, 4, 5, 6\)
Frequency: \(12, 13, 13, 15, 11, 16\)
Find the relative frequency, as a decimal, of
(i) a \(6\)
(ii) an even number
Solution
\(12 + 13 + 13\)\(=\)\(38\)
\(38 + 15 + 11 + 16\)\(=\)\(80\)
\(\dfrac{16}{80}\)\(=\)\(0.2\)
\(13 + 15 + 16\)\(=\)\(44\)
\(\dfrac{44}{80}\)\(=\)\(0.55\)
total 8016/80=0.213+15+16=44,44/80=0.55

(i) \(0.2\)
(ii) \(0.55\), close to the theoretical \(0.5\)

Example 3 — Two dice
Two dice were rolled \(60\) times and the sum recorded.
Sum: \(2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\)
Frequency: \(2, 3, 5, 6, 8, 10, 9, 7, 5, 3, 2\)
Find the experimental probability of
(i) a sum of \(10\)
(ii) an odd sum
(iii) a sum less than \(6\)
Solution
\(2 + 3 + 5 + 6 + 8 + 10\)\(=\)\(34\)
\(34 + 9 + 7 + 5 + 3 + 2\)\(=\)\(60\)
\(3 + 6 + 10 + 7 + 3\)\(=\)\(29\)
\(2 + 3 + 5 + 6\)\(=\)\(16\)
total 60 odd sums 3,5,7,9,113+6+10+7+3=29 less than 62+3+5+6=16

(i) \(\dfrac{5}{60} = \dfrac{1}{12}\)
(ii) odd sums are \(3, 5, 7, 9, 11\): \(\dfrac{29}{60}\)
(iii) sums \(2, 3, 4, 5\): \(\dfrac{16}{60} = \dfrac{4}{15}\)

Example 4 — Theory and experiment
A spinner has \(5\) equal sectors, \(3\) of them blue. It is spun \(50\) times and lands on blue \(34\) times. Find
(i) the theoretical probability of blue
(ii) the expected number of blues
(iii) the experimental probability of blue
Solution
\(\dfrac{3}{5} \times 50\)\(=\)\(30\)
\(\dfrac{34}{50}\)\(=\)\(\dfrac{17}{25}\)
3/5×50=30,34/50=17/25

(i) \(\dfrac{3}{5}\)
(ii) \(30\)
(iii) \(\dfrac{17}{25}\). A little more than expected; normal for \(50\) spins.

Example 5 — Estimating what is in a bag
A bag has \(20\) counters that are red, blue or green. Sara takes a counter, notes its colour and puts it back, \(60\) times. Red has a relative frequency of \(0.45\), blue came up \(15\) times and the rest were green.
(i) How many times did green come up?
(ii) Estimate the number of red counters.
Solution
\(0.45 \times 60\)\(=\)\(27\)
\(60 - 27 - 15\)\(=\)\(18\)
\(0.45 \times 20\)\(=\)\(9\)
red 0.45×60=27, green 60−27−15=18,0.45×20=9

(i) \(18\)
(ii) about \(9\) red counters

Example 6 — Predicting and bias
(i) A factory tests \(250\) batteries and finds \(4\) faulty. It makes \(8000\) batteries a week. About how many are faulty?
(ii) Leo rolls a die \(300\) times and gets \(95\) twos. Is the die fair?
Solution
\(\dfrac{4}{250} \times 8000\)\(=\)\(128\)
\(\dfrac{1}{6} \times 300\)\(=\)\(50\)
4/250×8000=1281/6×300=50

(i) about \(128\) faulty batteries
(ii) A fair die would give about \(50\) twos. \(95\) is far more, over many rolls (\(\dfrac{95}{300} \approx 0.32\), not about \(0.17\)), so the die is probably biased.

Common pitfalls

Dividing by the wrong number. \(12\) heads in \(40\) tosses is \(\dfrac{12}{40}\), not \(\dfrac{12}{28}\) (heads \(\div\) tails).
Forgetting to total the table. Add every frequency to get the number of trials before dividing.
Expecting exact results. Expected numbers are predictions; real results vary.
Calling a coin biased too soon. A few trials prove nothing. Look at many trials.

Frequently asked questions

What is experimental probability?

The number of times an event happened divided by the number of trials.

What is relative frequency?

Another name for experimental probability: frequency divided by the total frequency.

How do you find an expected number?

Multiply the probability by the number of trials. A fair coin tossed \(80\) times: about \(40\) heads.

Why is experimental probability different from theoretical probability?

Chance results vary. With more trials they usually get closer.

How can you tell if a die is biased?

After many rolls, one number comes up far more or far less often than expected.