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AI HL · 13.1

Probability

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13.1.1

Theoretical and experimental probability

13.1.1

Theoretical and experimental probability

Target: I can find a theoretical probability by counting favourable outcomes and dividing by all equally likely outcomes. (+3 more)

By the end of this section

Key points

Theoretical and experimental probability

  • Probability is a number from 0 (impossible) to 1 (certain). Give it as a fraction, a decimal or a percentage.
  • If all outcomes are equally likely: \(P(A)=\frac{n(A)}{n(U)}\), favourable outcomes over all outcomes.
  • Count n(A) by listing the outcomes. List them, do not guess.
  • Experimental probability (relative frequency) is the number of times the event happened ÷ the number of trials. It is an estimate: a larger sample gives a better one.

Complementary events

  • The complement \(A'\) is "A does not happen". \(P(A')=1-P(A)\).
  • "At least one" is the complement of "none". Find P(none), then take it away from 1.
  • For repeated independent trials, P(none) = (P(not A)) to the power of the number of trials.

Expected number

  • Expected number of occurrences = P(A) × number of trials.
  • The answer need not be a whole number. It is a long-run average, not a prediction for one run.
  • Estimate a population count by using a sample's relative frequency as P.

Explore: relative frequency

Choose the probability, then add trials. The line is the relative frequency so far. The dashed line is the probability it settles towards.

What is being repeated

Trials, n
Heads so far
Expected, n × p
Count − expected

Press "10 more" to start.

  1. Press "1 more" ten times. How far from p can the relative frequency be after only 10 trials?
  2. Keep pressing "100 more". What happens to the line? Does it ever stop moving?
  3. Compare the count with n × p. Is it exactly equal? Should it be?
  4. Press "New run" with the same p. Is the start of the line the same? Is the end?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A probability from equally likely outcomes

A fair spinner is numbered 1 to 12. Find the probability that it lands on a square number. Give your answer as a decimal.

Working
  1. Sample space

    n(U) = 12

    The spinner is fair, so the 12 outcomes are equally likely.

  2. Favourable outcomes

    Square numbers: 1, 4, 9. n(A) = 3

    List them. Do not miss 1.

  3. Probability

    \(P(A)=\frac{n(A)}{n(U)}=\frac{1}{4}\) = 0.25

    Favourable outcomes over all outcomes.

Video walkthrough coming
At least one, from P(none)

A courier delivers several parcels to a shop. The probability that none of the parcels is damaged is 0.72. Find the probability that at least one parcel is damaged.

Working
  1. Name the events

    A = at least one damaged. A′ = none damaged.

    These two events cover every possibility and cannot both happen.

  2. Complement

    \(P(A)=1-P(A')=1-0.72=0.28\)

    Probabilities of an event and its complement add to 1.

Video walkthrough coming
At least one head in three tosses

A fair coin is tossed 3 times. Find the probability of at least one head. Give your answer as a decimal.

Working
  1. The complement

    At least one head is the opposite of 'no heads' (TTT).

    Counting 'at least one' directly means three cases. The complement is one case.

  2. P(none)

    \(P(TTT)=\frac12\times\frac12\times\frac12=\frac{1}{8}\)

    Multiply the three tosses.

  3. Subtract

    \(P(\text{at least one head})=1-\frac{1}{8}=\frac{7}{8}\) = 0.875

    Complement rule.

Video walkthrough coming
Expected number of occurrences

A goalkeeper saves a penalty with probability 0.45. She faces 40 penalties in a season. Find the expected number of saves.

Working
  1. Use the rule

    Expected number = P(save) × number of trials

    This is the long-run average, not a promise.

  2. Calculate

    0.45 × 40 = 18

    A whole number here. If it is not, leave it as a decimal.

Video walkthrough coming
Estimating from a sample

A random sample of 80 packets of biscuits is checked. 6 packets contain a broken biscuit. (a) Estimate the probability that a packet contains a broken biscuit. (b) A factory makes 1500 packets in a day. Estimate how many contain a broken biscuit.

Working
  1. Relative frequency

    \(\frac{6}{80}=\frac{3}{40}\) = 0.075

    Probability estimated = number with the property ÷ sample size.

  2. Scale up

    0.075 × 1500 = 112.5

    Use the estimate as P in the expected-number rule.

  3. Answer (b)

    About 113 packets

    It is an estimate, so round to a sensible whole number.

Video walkthrough coming