Probability: experiments, Venn and tree diagrams – Year 11 GCSE maths revision notes to download as PDF

Worked example 1: a spinner

A spinner is red with probability 0.3 and blue with probability 0.5. Find \(P(\text{green})\).

\(1-0.3-0.5=0.2\).

Worked example 2: drawing pins and fairness

320 out of 500 drawing pins land point up. Estimate the probability and the expected number in 150 drops. A spinner lands on red 28 times in 80 spins; should red have probability 0.25?

The estimate is \(\dfrac{320}{500}=0.64\), so in 150 drops we expect \(0.64\times150=96\). For the spinner, the relative frequency is \(\dfrac{28}{80}=0.35\) and we would expect \(0.25\times80=20\) reds. The spinner may be biased, but 80 trials is not many, so do more trials before deciding. Give the numbers, then a conclusion.

Worked example 3: dice and coins

Two dice are added: find \(P(7)\) and \(P(8)\). Two coins are tossed: find \(P(\text{two heads})\) and \(P(\text{at least one head})\).

A sum of 7 occurs in 6 ways, so \(P(7)=\dfrac6{36}=\dfrac16\). A sum of 8 occurs in 5 ways \((2,6),(3,5),(4,4),(5,3),(6,2)\), so \(P(8)=\dfrac5{36}\). For the coins, \(P(\text{two heads})=\dfrac14\) and \(P(\text{at least one head})=\dfrac34\).

Worked example 4: football and tennis

30 students: 18 play football (F), 12 play tennis (T), 5 play both. Find \(P(F\cup T)\), \(P(F\cap T)\) and \(P(F\text{ only})\).

Both \(=5\). F only \(=13\). T only \(=7\). Neither \(=30-25=5\). Then \(P(F\cup T)=\dfrac{25}{30}=\dfrac56\), \(P(F\cap T)=\dfrac{5}{30}=\dfrac16\) and \(P(F\text{ only})=\dfrac{13}{30}\).

Worked example 5: with replacement

A bag has 5 red and 3 blue counters. Two are taken with replacement. Find \(P(RR)\) and \(P(\text{one of each})\).

\(P(RR)=\dfrac58\times\dfrac58=\dfrac{25}{64}\). \(P(\text{one of each})=2\times\dfrac58\times\dfrac38=\dfrac{30}{64}=\dfrac{15}{32}\).

Worked example 6: without replacement

Same bag, but the first counter is not put back. Find \(P(RR)\), \(P(BB)\), \(P(\text{different})\) and \(P(\text{at least one blue})\).

After one is removed there are 7 left. \(P(RR)=\dfrac58\times\dfrac47=\dfrac5{14}\). \(P(BB)=\dfrac38\times\dfrac27=\dfrac3{28}\). \(P(\text{different})=1-\dfrac5{14}-\dfrac3{28}=\dfrac{15}{28}\). \(P(\text{at least one blue})=1-\dfrac5{14}=\dfrac9{14}\).

Worked example 7: Using relative frequency to predict

A bag contains red and blue counters. In 300 trials a red counter was taken 123 times. Estimate the probability of red and the number of red counters in a bag of 50.

The relative frequency of red is \(\dfrac{123}{300}=0.41\). This is the best estimate of \(P(\text{red})\). In a bag of 50 counters we expect about \(50\times0.41=20.5\), so about 20 or 21 red counters. This is an estimate: with more trials it would be more reliable.

Worked example 8: The addition rule for mutually exclusive events

A card is chosen from a pack where \(P(\text{red})=0.3\), \(P(\text{blue})=0.5\) and the rest are green. Find \(P(\text{red or blue})\) and \(P(\text{green})\).

The events are mutually exclusive because a card cannot be two colours at once. So \(P(\text{red or blue})=0.3+0.5=0.8\). The probabilities add to 1, so \(P(\text{green})=1-0.8=0.2\). The word "or" usually means add and the word "and" usually means multiply.

Worked example 9: The general addition rule with a Venn diagram

\(P(A)=0.5\), \(P(B)=0.4\) and \(P(A\cap B)=0.15\). Find \(P(A\cup B)\) and \(P(\text{neither})\).

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.4-0.15=0.75\). We subtract the overlap because it was counted twice. \(P(\text{neither})=1-0.75=0.25\). On a Venn diagram: \(A\) only 0.35, both 0.15, \(B\) only 0.25, neither 0.25.

Worked example 10: A tree diagram with three branches

A fair spinner is red, blue or green with equal probability. It is spun twice. Find the probability of getting two different colours.

The probability of getting the same colour twice is \(\dfrac13\times\dfrac13=\dfrac19\) for each colour, so \(3\times\dfrac19=\dfrac13\). Therefore different colours has probability \(1-\dfrac13=\dfrac23\). Using the complement is quicker than listing all six different pairs.

Worked example 11: Expected frequency

The probability that a train is late is 0.15. How many late trains are expected in 40 journeys? Is it possible for none to be late?

Expected number \(=40\times0.15=6\). This is an average and the real number varies. It is possible for none to be late: the probability is \(0.85^{40}=0.0015\), which is very small but not zero. Expected value does not mean "exactly".

Worked example 12: Sample space for a coin and a die

A fair coin is tossed and a fair die is rolled. Find the probability of getting heads and an even number.

There are \(2\times6=12\) equally likely outcomes. Heads and even: (H,2), (H,4), (H,6), which is 3 outcomes. The probability is \(\dfrac3{12}=\dfrac14\). Because the events are independent, you could also multiply: \(\dfrac12\times\dfrac12=\dfrac14\).

Worked example 13: Fairness and bias

A spinner is spun 80 times and lands on red 30 times. Is it fair if it is supposed to land on red with probability \(\tfrac13\)?

A fair spinner would give about \(80\times\dfrac13=26.7\) reds. The experiment gave 30, which is 3.3 more. This is a small difference for 80 trials, so there is no strong evidence that the spinner is biased. More trials would give a clearer answer.