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ScotlandApplications of MathematicsSyllabus dot point

How do you calculate the probability of an event, interpret it on the likelihood scale, and use probability and risk to make and justify a decision?

Calculating the probability of an event as a fraction, decimal or percentage, interpreting probability on the scale from 0 to 1, using expected frequency, and using probability and risk to make and justify decisions.

A focused answer to the SQA National 5 Applications of Mathematics statistics content on probability, covering calculating the probability of an event, the likelihood scale from 0 to 1, expected frequency, and using probability and risk to make and justify decisions.

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  1. What this dot point is asking
  2. Calculating probability
  3. The likelihood scale and complementary events
  4. Expected frequency, risk and decisions
  5. Examples in context
  6. Try this

What this dot point is asking

The SQA wants you to calculate the probability of an event as a fraction, decimal or percentage, place it on the likelihood scale from 00 to 11, work out an expected frequency, and use probability and risk to make and justify a decision.

Calculating probability

Probability measures the chance of an event happening. When all outcomes are equally likely, it is a simple fraction of favourable outcomes over the total.

The likelihood scale and complementary events

Every probability sits on a scale from 00 to 11. An impossible event has probability 0,0, a certain event 11, and an even chance 0.50.5.

Expected frequency, risk and decisions

Expected frequency predicts how many times an event should happen over many trials. It turns a probability into a count, which helps weigh up risk.

Risk questions ask you to use a probability to judge how safe or worthwhile something is, and to justify the decision. A low probability of a serious fault may still matter if the consequences are severe, so the decision must be explained, not just stated.

Probability can be found from data as well as from equally likely outcomes. If a bus was late on 1515 of the last 6060 days, the experimental probability of it being late is 1560=14\dfrac{15}{60} = \dfrac{1}{4}, which can then be used to predict future lateness. The more data you have, the more reliable this estimate becomes, which is why expected frequency over many trials is trusted more than a single result.

Comparing two options on risk uses the same idea: the option with the lower probability of a bad outcome, or the lower expected number of failures, is usually the safer choice, but the size of the consequence matters too.

Examples in context

Probability and risk underpin everyday choices: a weather forecast gives the chance of rain, a manufacturer estimates how many products will be faulty, an insurer prices a policy on the risk of a claim. The SQA tests calculating the probability, finding an expected frequency, and using these to make and justify a decision, the skills here.

Try this

Q1. A fair coin is tossed. Find the probability of heads. [1 mark]

  • Cue. 12\dfrac{1}{2}.

Q2. P(late)=0.15P(\text{late}) = 0.15. Find the probability of not being late. [2 marks]

  • Cue. 10.15=0.851 - 0.15 = 0.85.

Q3. P(prize)=0.1P(\text{prize}) = 0.1 and 200200 tickets are bought. Find the expected number of prizes. [2 marks]

  • Cue. 0.1×200=200.1 \times 200 = 20.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA N5 Apps style3 marksA bag holds 55 red, 33 blue and 22 green counters. One counter is taken at random. Find the probability it is red, and the probability it is not green.
Show worked answer →

There are 5+3+2=105 + 3 + 2 = 10 counters in total (1 mark for the total). The probability of red is the number of red over the total: P(red)=510=12P(\text{red}) = \dfrac{5}{10} = \dfrac{1}{2} (1 mark). The probability of not green is the 88 non-green counters over the total: P(not green)=810=45P(\text{not green}) = \dfrac{8}{10} = \dfrac{4}{5} (1 mark). Markers reward the total, the red probability, and the not-green probability. The not-green probability also equals 1P(green)1 - P(\text{green}).

SQA N5 Apps style3 marksThe probability that a machine produces a faulty part is 0.040.04. In a batch of 15001500 parts, how many are expected to be faulty, and what does this suggest about the machine?
Show worked answer →

The expected number is the probability multiplied by the number of trials: 0.04×15000.04 \times 1500 (1 mark). Evaluate: 0.04×1500=600.04 \times 1500 = 60 parts expected to be faulty (1 mark). This is a low fault rate of 4%4\%, so the machine is fairly reliable, though 6060 faulty parts in a batch may still need checking (1 mark). Markers reward the expected-frequency calculation, the value, and a justified comment in context.

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