Probabilities Summing to 1

Each time a dice is rolled or a spinner is spun, exactly one of the possible outcomes happens. So the probabilities of all the outcomes add up to 1. On a fair dice each number has probability \frac{1}{6}, and six lots of \frac{1}{6} make 1.

A biased dice or spinner is not fair, so its outcomes are not equally likely. You cannot count the sides or sectors. Instead you use the probabilities you are given, and the fact that they add up to 1.

In the exam you are usually asked to work out the probability that something does not happen, or to find the missing probability in a table, for 1 or 2 marks. Harder questions, for 4 or 5 marks, give two missing probabilities linked by a fact, then ask how many counters or sweets there are.

A bag contains only red, blue and green counters. One counter is taken at random. It must be red, blue or green, and it cannot be two colours at once. So

P(\text{red})+P(\text{blue})+P(\text{green})=1

The word only in the question tells you there are no other colours.

An event either happens or it does not, so these two probabilities add up to 1 as well:

P(\text{not }A)=1-P(A)

where A is the event. If P(\text{win})=\frac{4}{9}, then P(\text{not win})=\frac{9}{9}-\frac{4}{9}=\frac{5}{9}

Outcomes that cannot happen at the same time are called mutually exclusive. You can add their probabilities: P(\text{blue or green})=P(\text{blue})+P(\text{green}). Only add when the outcomes cannot both happen. A student can own a dog and a cat, so those two probabilities should not just be added.

The total also shows when a list of probabilities cannot be right. If the probabilities add up to more than 1, or any one of them is less than 0 or more than 1, there is a mistake.

A probability table lists every outcome with its probability. When one probability is missing, add the probabilities you know, then subtract the total from 1.

Line up the decimal points when you add and subtract. If the known probabilities add up to 0.81, the missing one is 1.00-0.81=0.19, not 0.29 or 1.19.

Check your answer: every probability in the table, including yours, should add up to exactly 1.

Some tables mix fractions and decimals. Change them all into the same form before you add. Decimals are usually easiest, so learn these:

• \frac{1}{2}=0.5
• \frac{1}{4}=0.25 and \frac{3}{4}=0.75
• \frac{1}{5}=0.2, so \frac{2}{5}=0.4
• \frac{1}{10}=0.1, so \frac{3}{10}=0.3
• \frac{1}{20}=0.05

If every probability is a fraction, use a common denominator instead. With \frac{1}{3} and \frac{1}{4}, use twelfths: \frac{4}{12}+\frac{3}{12}=\frac{7}{12}, so the missing probability is \frac{12}{12}-\frac{7}{12}=\frac{5}{12}

Give your answer in the form the question asks for. If it says "Give your answer as a fraction", change a decimal such as 0.35 into \frac{35}{100}=\frac{7}{20}

Sometimes two probabilities are missing. First work out what is left for the two of them: 1 minus the probabilities you know. Then share what is left using the extra fact in the question.

• The same (or equally likely): share it equally. If 0.3 is left, each gets 0.15.
• Twice as likely: make 3 equal parts, 1 part for one outcome and 2 parts for the other. If 0.48 is left, one part is 0.48\div3=0.16, so the two are 0.16 and 0.32.
• In the ratio 3:5: make 3+5=8 equal parts. If 0.4 is left, one part is 0.4\div8=0.05, so the two are 0.15 and 0.25.
• 0.1 more than: take off the extra first. If 0.5 is left, 0.5-0.1=0.4 is shared equally, so the two are 0.2 and 0.3.

Check your two answers: they must add up to what was left, and they must fit the fact in the question.

A probability tells you what fraction of the things are of each kind. To find how many there are, multiply the probability by the total:

\text{number}=\text{probability}\times\text{total}

With 60 counters and P(\text{blue})=0.35, there are 0.35\times60=21 blue counters.

To find the total, divide the number you know by its probability:

\text{total}=\text{number}\div\text{probability}

If P(\text{green})=0.4 and there are 26 green counters, the total is 26\div0.4=65 counters.

Without a calculator, work in equal parts. 0.4 stands for 26 counters, so 0.2 stands for 13. The whole bag, 1, is 5 lots of 0.2, so it holds 5\times13=65 counters.

The same multiplying gives an estimate of how often something will happen. If P(\text{tails})=0.42 and a coin is thrown 50 times, expect about 0.42\times50=21 tails.

• Writing any two numbers that add up to what is left. The two missing probabilities must also fit the fact in the question: equal, twice as likely, a ratio or a set amount more. Never write all of what is left in both boxes.
• Dividing by each number in a ratio separately. For a ratio of 3:5, add the parts, 3+5=8, and divide what is left by 8.
• Giving the shares to the wrong outcomes. The outcome that is more likely gets the bigger share, so check which one the question names.
• Multiplying instead of dividing to find the total. If 0.4 of the counters is 26 counters, the total is 26\div0.4=65; the total must be bigger than the part.
• Adding on instead of scaling up. If 0.2 stands for 13 counters, then 0.4 stands for 26 counters, not 15.
• Ending with probabilities that do not add up to 1. Add every probability in the table, including yours, and check the total is exactly 1, with none bigger than 1.