Calculating probabilities
🎯 In this topic you will
- Find the probability of complementary events
- Use lists and diagrams to show equally likely outcomes
- Use lists and diagrams of outcomes to calculate probabilities
🧠 Key Word
- complementary event
Show Definition
- complementary event: In probability, the event that represents all outcomes not included in the original event. For any event, its probability plus the probability of its complementary event equals $1$.
🎡 Complementary Events with a Spinner
This is a spinner.
The probability that it points to red is $0.2$.
The probability that it points to blue is $0.15$.
We can write those probabilities as $P(\text{red}) = 0.2$ and $P(\text{blue}) = 0.15$.
The sum of the probabilities for all six colours is $1$.
This means the probability the spinner does not point to red,
$P(\text{not red}) = 1 - 0.2 = 0.8$
The probability the spinner does not point to blue,
$P(\text{not blue}) = 1 - 0.15 = 0.85$
Getting blue and not getting blue are complementary events.
One of them must happen and they cannot both happen.
If $A$ is an event and $A’$ is the complementary event, then
$P(A’) = 1 - P(A)$
❓ EXERCISES
1. The probability that a football team will win a match is $0.3$.
The probability that the team will draw is $0.1$.
Work out the probability that the team will
a. not win b. not draw c. lose d. not lose.
👀 Show answer
a. $1 - 0.3 = 0.7$
b. $1 - 0.1 = 0.9$
c. $1 - (0.3 + 0.1) = 0.6$
d. $1 - 0.6 = 0.4$
2. Tomorrow must be hotter, colder or the same temperature as today.
The probability it will be hotter is $55\%$.
The probability it will be colder is $25\%$.
Work out the probability that it will
a. not be hotter b. not be colder c. not be the same temperature.
👀 Show answer
Total probability = $100\%$.
Hotter = $55\%$, Colder = $25\%$, Same = $20\%$.
a. $100\% - 55\% = 45\%$
b. $100\% - 25\% = 75\%$
c. $100\% - 20\% = 80\%$
3. A spinner has five colours on it.
The probability it shows green is $0.32$.
The probability it shows purple is $0.17$.
Find the probability that the colour is
a. not green b. not purple.
👀 Show answer
a. $1 - 0.32 = 0.68$
b. $1 - 0.17 = 0.83$
❓ EXERCISES
4. There are lots of coloured toys in a box. Here are the percentages of some of the colours.
| Colour | yellow | orange | red | green |
|---|---|---|---|---|
| Percentage | $15\%$ | $25\%$ | $30\%$ | $10\%$ |
a. Why do the percentages add up to less than $100\%$?
A child takes a toy at random.
b. Find the probability that the toy is
i not orange ii not green iii not red iv not yellow.
👀 Show answer
a. The table lists only some colours; the missing $20\%$ belongs to other colours.
b. Using complements:
- i.$1-0.25=0.75$
- ii.$1-0.10=0.90$
- iii.$1-0.30=0.70$
- iv.$1-0.15=0.85$
5. Two dice are thrown. Find the probability that
a. both dice show $5$ b. one dice shows a $5$ and the other does not c. neither dice shows a $5$.
👀 Show answer
Total outcomes $=6\times6=36$.
a.$\tfrac{1}{36}$
b. Exactly one $5$: $10$ outcomes → $\tfrac{10}{36}=\tfrac{5}{18}$
c. Neither $5$: $5\times5=25$ outcomes → $\tfrac{25}{36}$
6. Two dice are thrown. The numbers are added together.
a. Draw a table to show all the possible outcomes.
b. Find the probability that the total is
i $3$ ii $7$ iii $12$ iv $9$
c. Copy and complete this table of probabilities.
| Total | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | $\tfrac{1}{36}$ | $\tfrac{2}{36}=\tfrac{1}{18}$ | $\tfrac{3}{36}=\tfrac{1}{12}$ | $\tfrac{4}{36}=\tfrac{1}{9}$ | $\tfrac{5}{36}$ | $\tfrac{6}{36}=\tfrac{1}{6}$ | $\tfrac{5}{36}$ | $\tfrac{4}{36}=\tfrac{1}{9}$ | $\tfrac{3}{36}=\tfrac{1}{12}$ | $\tfrac{2}{36}=\tfrac{1}{18}$ | $\tfrac{1}{36}$ |
👀 Show answer
b. From the distribution: i$\tfrac{2}{36}=\tfrac{1}{18}$, ii$\tfrac{6}{36}=\tfrac{1}{6}$, iii$\tfrac{1}{36}$, iv$\tfrac{4}{36}=\tfrac{1}{9}$.
c. Table completed above (counts across sums: $1,2,3,4,5,6,5,4,3,2,1$ divided by $36$).
7. Two dice are thrown. The numbers are added together.
a. Find the probability that the total is
i $5$ or less ii more than $5$ iii $10$ or more iv less than $10$ v a prime number.
b. Find an event with a probability of $\tfrac{7}{36}$.
c. Give your answer to part b. to a partner to check it is correct.
👀 Show answer
a. Using the table in Q6:
- i.$\dfrac{1+2+3+4}{36}=\dfrac{10}{36}=\dfrac{5}{18}$
- ii.$1-\dfrac{10}{36}=\dfrac{26}{36}=\dfrac{13}{18}$
- iii.$\dfrac{3+2+1}{36}=\dfrac{6}{36}=\dfrac{1}{6}$
- iv.$1-\dfrac{6}{36}=\dfrac{30}{36}=\dfrac{5}{6}$
- v. Prime totals $2,3,5,7,11$ → counts $1+2+4+6+2=15$ → $\dfrac{15}{36}=\dfrac{5}{12}$
b. Example: “the total is $4$ or $9$” (counts $3+4=7$) → $\tfrac{7}{36}$.
c. Check by verifying the counts on the Q6 table.
❓ EXERCISES
8. A fair coin and a fair dice are thrown. This table shows the possible outcomes.
| Coin | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|---|---|---|---|---|---|
| H | H1 | H2 | H3 | H4 | H5 | H6 |
| T | T1 | T2 | T3 | T4 | T5 | T6 |
a. Copy and complete the table.
b. How many outcomes are there? Are they all equally likely?
c. Find the probability of
i. $6$ and a tail ii. $4$ and a head iii. a head and an even number iv. a tail and a number less than $3$.
d. Find the probability of each of the events in part c not happening.
e. Describe an event with a probability of $\tfrac{5}{12}$.
f. Give your answer to part e to a partner to check.
👀 Show answer
a. Table completed above.
b. $12$ outcomes. Yes, each is equally likely.
c.
- i. $1/12$
- ii. $1/12$
- iii. $3/12=1/4$
- iv. $2/12=1/6$
d. Complements: $11/12,\;11/12,\;9/12=3/4,\;10/12=5/6$.
e. Example: “Tail and dice shows 4, 5, or 6” → 3 favourable out of 12 → $3/12=1/4$ (not equal to $5/12$). Correct example: “Head with odd dice OR Tail with dice 1 or 2” → $5$ favourable → $5/12$.
9. Here are two spinners.
a. The two spinners are spun. Draw a diagram to show all the outcomes.
b. Work out the probability that
i. both spinners show a $1$ ii. neither spinner shows a $1$ iii. both spinners show the same number iv. the spinners do not show the same number.
c. The two scores are added together. Draw a table to show the possible totals.
d. Find the probability that the total is
i. $4$ ii. $5$ iii. not $7$ iv. a multiple of $3$.
e. Now the scores on the spinners are multiplied. Draw a table to show the possible products.
f. Find the probability of each of the different possible products.
g. Find the probability that the product is
i. $6$ or more ii. less than $6$ iii. an odd number iv. an even number.
👀 Show answer
b. i. $1/16$ ii. $9/16$ iii. $4/16=1/4$ iv. $12/16=3/4$
d. i. $1/16$ ii. $2/16=1/8$ iii. $15/16$ iv. $6/16=3/8$
f. Product distribution requires table; probabilities follow counts over 16.
g. i. $11/16$ ii. $5/16$ iii. $8/16=1/2$ iv. $8/16=1/2$
❓ EXERCISES
10.
a. Two fair coins are flipped. Copy and complete this table to show the outcomes.
| Second coin H | Second coin T | |
|---|---|---|
| First coin H | HH | HT |
| First coin T | TH | TT |
b. Read what Arun says. Explain why Arun is not correct.
c. Find the probability of
i $2$ heads ii $2$ tails iii a head and a tail.
d. Another way to show the outcomes when two fair coins are thrown is a tree diagram. Copy the tree diagram and fill in the missing outcomes.
e. Explain how the table in part a. and the tree diagram in part d. show the same outcomes.
f. Three fair coins are thrown. One possible outcome is HHH, a head on all three coins. List all the possible outcomes in this way.
g. Draw a tree diagram to show the results of throwing three fair coins. Use it to check your answer to part f.
h. When three fair coins are thrown, find the probability of
i $3$ heads ii $3$ tails iii not getting $3$ heads iv $2$ heads and $1$ tail v $1$ head and $2$ tails.
👀 Show answer
a. Completed table shown above: HH, HT, TH, TT.
b. Arun grouped HT and TH as one outcome. They are two distinct equally likely outcomes, so there are $4$ outcomes, not $3$. Thus $P(2\ \text{heads})=\tfrac{1}{4}$, not $\tfrac{1}{3}$.
c. i $\tfrac{1}{4}$ ii $\tfrac{1}{4}$ iii $\tfrac{2}{4}=\tfrac{1}{2}$
d. Tree ends: HH, HT, TH, TT (each with probability $\tfrac{1}{4}$).
e. Both methods list the same sample space {HH, HT, TH, TT}; the table arranges by row/column, the tree arranges by stages.
f. Outcomes for three coins: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
g. Three-level tree confirms $8$ equally likely outcomes.
h. i $\tfrac{1}{8}$ ii $\tfrac{1}{8}$ iii $1-\tfrac{1}{8}=\tfrac{7}{8}$ iv $\tfrac{3}{8}$ v $\tfrac{3}{8}$
🧠 Think like a Mathematician
Investigation: Investigate the possible outcomes when 4 fair coins are thrown.
You should find all the possible outcomes and calculate probabilities of different events. Use your experience from Question 10 to help you.
Method:
- List all possible outcomes for four coin flips (e.g., HHHH, HHHT, ... , TTTT).
- Count the total number of outcomes.
- Group the outcomes by the number of heads (0, 1, 2, 3, 4).
- Calculate the probability of each group by dividing the count by the total.
Follow-up Questions:
Show Answers
- 1: There are $2^4 = 16$ total outcomes.
- 2: Number of ways to get 2 heads = $\binom{4}{2} = 6$. So probability = $\tfrac{6}{16} = \tfrac{3}{8}$.
- 3: Outcomes with at least 3 heads = $\binom{4}{3} + \binom{4}{4} = 4+1=5$. Probability = $\tfrac{5}{16}$.
- 4: The probabilities follow the binomial pattern. With more coins, the distribution of heads becomes closer to a “bell” shape (binomial distribution).
❓ EXERCISES
12. Zara has three cards with numbers on them: $2$, $4$, $5$.
She puts the cards side by side in a random order to make a $3$-digit number.
a. List all the possible numbers. Make sure you have found them all.
b. Find the probability that the number formed is
i an odd number ii an even number iii more than $400$.
Zara adds an extra card. Now she has four cards: $2$, $4$, $5$, $8$.
Zara takes two cards at random and places them side by side to make a $2$-digit number.
c. List all the possible numbers she can make. Make sure you have found them all.
d. Find the probability that the $2$-digit number
i is $48$ ii is not $48$ iii is an odd number iv is an even number v includes the digit $2$.
Now Zara takes three cards at random and places them side by side to make a $3$-digit number.
e. List all the possible numbers she can make.
f. Find the probability that the $3$-digit number is
i an odd number ii an even number iii less than $500$.
👀 Show answer
a. Permutations of $\{2,4,5\}$ (total $3!=6$): $245,254,425,452,524,542$.
b. Sample space size $=6$.
i. Odd ⇒ last digit $5$: $245,425$ → $\tfrac{2}{6}=\tfrac{1}{3}$.
ii. Even ⇒ last digit $2$ or $4$: $254,524,452,542$ → $\tfrac{4}{6}=\tfrac{2}{3}$.
iii. More than $400$ ⇒ hundreds digit $4$ or $5$: $425,452,524,542$ → $\tfrac{4}{6}=\tfrac{2}{3}$.
c. Two-digit numbers from $\{2,4,5,8\}$ without repetition (total $4P_2=12$):
$24,25,28,42,45,48,52,54,58,82,84,85$.
d. Denominator $=12$.
i.$48$ only → $\tfrac{1}{12}$.
ii. Not $48$ → $\tfrac{11}{12}$.
iii. Odd ⇒ units $5$: $25,45,85$ → $\tfrac{3}{12}=\tfrac{1}{4}$.
iv. Even ⇒ units in $\{2,4,8\}$: $9$ cases → $\tfrac{9}{12}=\tfrac{3}{4}$.
v. Includes digit $2$: $\{24,25,28,42,52,82\}$ → $\tfrac{6}{12}=\tfrac{1}{2}$.
e. Three-digit numbers from $\{2,4,5,8\}$ without repetition (total $4P_3=24$):
$245,254,248,284,258,285,$
$425,452,428,482,458,485,$
$524,542,528,582,548,584,$
$824,842,825,852,845,854$.
f. Denominator $=24$.
i. Odd ⇒ units $5$; first two are permutations of two from $\{2,4,8\}$: $3\times2=6$ cases → $\tfrac{6}{24}=\tfrac{1}{4}$.
ii. Even ⇒ units in $\{2,4,8\}$: $3\times(3\times2)=18$ cases → $\tfrac{18}{24}=\tfrac{3}{4}$.
iii. Less than $500$ ⇒ hundreds digit $2$ or $4$: $6+6=12$ cases → $\tfrac{12}{24}=\tfrac{1}{2}$.