13.3 Conditional Probability & Expected Frequency

Papers I and J (Years 11-12) push probability into its most realistic territory: questions where information changes the odds. Medical screening, quality control, survey data — all of these are conditional probability problems, and ICAS presents them through two-way tables and tree diagrams. The same papers also expect you to connect probability to data through expected frequency and relative frequency. Neither idea needs new arithmetic; both need careful reading of what the denominator should be.

Conditional Probability: Information Changes Everything

P(A|B) is read 'the probability of A given B'. It means: restrict your attention to the situations where B has happened, then find the fraction of those in which A also happens.

Formally: P(A|B) = P(A and B) / P(B)

You rarely need the formula explicitly at this level — a table does the work.

Worked example 1: a two-way table

100 students were asked whether they play a sport and whether they learn a musical instrument:

A student is chosen at random.

• P(instrument) = 40/100 = 2/5 — the grand total is the denominator.
• P(instrument | plays sport): we are told the student plays sport, so only the 60 students in the sport row count. Of those, 24 learn an instrument: 24/60 = 2/5.
• P(sport | instrument): now the condition is the instrument column: 24 out of 40, so 3/5.

Notice the last two answers differ. P(A|B) and P(B|A) are generally not equal — the 'given' condition sets the denominator, and swapping it changes the question. ICAS tests precisely this distinction.

Worked example 2: reading a tree conditionally

A box has 4 red and 6 black balls; two are drawn without replacement. The second-stage branches are already conditional: after 'red first', P(black second | red first) = 6/9, because 9 balls remain and all 6 black ones are still there. Whenever a tree branch probability has a reduced denominator, you are looking at a conditional probability wearing a disguise.

Expected Frequency

Probability predicts the long run. If an event has probability p and you repeat the experiment n times, the expected number of occurrences is:

Expected frequency = p x n

Worked example 3

A fair die is rolled 300 times. How many fives would you expect?

Expected frequency = 1/6 x 300 = 50.

'Expect' does not mean 'guarantee'. Getting 47 or 54 fives is perfectly normal; expected frequency is the average over many repetitions. If an ICAS option says 'exactly 50 every time', it is wrong — the prediction is a long-run average.

Worked example 4: reverse problems

A biased coin is tossed 500 times and lands heads 320 times. Estimate P(head) and predict heads in the next 200 tosses.

• Estimated P(head) = 320/500 = 0.64
• Expected heads in 200 tosses = 0.64 x 200 = 128

Experimental versus Theoretical Probability

• Theoretical probability comes from reasoning about equally likely outcomes: P(6 on a fair die) = 1/6.
• Experimental probability (relative frequency) comes from doing it: if a drawing pin lands point-up 37 times in 100 drops, the relative frequency is 37/100 = 0.37, and that is your best estimate of its probability. There is no symmetry argument for a drawing pin, so experiment is the only route.

Two facts ICAS likes to test:

1. More trials give a more reliable estimate. 0.37 from 100 drops is less trustworthy than 0.34 from 10,000 drops. If asked whose estimate is best, choose the person with the most trials — not the answer closest to a 'nice' fraction.
2. A mismatch between relative frequency and theoretical probability suggests bias. If a die shows six 90 times in 300 rolls (relative frequency 0.3 versus the theoretical 1/6 ≈ 0.167), the die is probably biased — the gap is far larger than ordinary variation.

Worked example 5: combining the ideas

A factory machine produces a defective part with probability 0.02. In a batch of 4,500 parts, how many defectives are expected?

Expected frequency = 0.02 x 4,500 = 90.

Follow-up ICAS would ask: if 140 defectives are actually found, what does that suggest? The relative frequency is 140/4,500 ≈ 0.031, well above 0.02, so the machine's defect rate has likely increased — the claim of 0.02 is in doubt.

Traps at Senior-Paper Level

• Wrong denominator in 'given' questions. Underline the condition and use its row or column total.
• Confusing P(A and B) with P(A|B). From the table above, P(sport and instrument) = 24/100, but P(instrument | sport) = 24/60.
• Treating expected frequency as exact. It is a prediction about the average, and real results vary around it.
• Judging an estimate by its looks. Reliability comes from the number of trials, not from how neat the fraction is.