How does the denominator change without replacement?

One ball leaves the bag, so both the numerator and the denominator fall by one.

How is a conditional probability found?

Replace the denominator with the group named by the condition, such as the nine balls left after a red is drawn.

S4 Mathematics: Probability (Paper 4)

2026-10-31

Tutor Prep Guide · Secondary Mathematics

Ten original questions practise listing and conditional probability: twenty balls give a quarter for red, a die gives one half for an even number, four aces in fifty-two cards give one thirteenth and a spinner from one to ten gives one tenth for a multiple of seven. Later items count the five ways to total eight with two dice, use the one third for a second red given a first red, combine independent events at 0.6 and 0.5 into 0.3 and 0.8, expect twenty sixes in one hundred and twenty rolls, take the complement for at least one red without replacement, and repair the error of treating draws without replacement as independent.

Direct answer　Write down the number of equally likely outcomes and count the favourable ones; without replacement both the numerator and the denominator fall on the second draw.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Mathematics: Probability
Level	Four basic, three advanced, two challenge and one error-repair item

02　Questions (10 items)

FoundationA bag holds 5 red, 7 white and 8 blue balls. One ball is drawn at random. Find the probability that it is red.
FoundationA fair die is rolled once. Find the probability of getting an even number.
FoundationOne card is drawn from a standard pack of 52. Find the probability that it is an ace.
FoundationA fair spinner has faces numbered 1 to 10. Find the probability that the number is a multiple of 7.
IntermediateTwo fair dice are rolled together. Find the probability that the total is 8.
IntermediateA bag holds 4 red and 6 white balls and two balls are drawn without replacement. Given that the first ball is red, find the probability that the second is also red.
IntermediateP(A) = 0.6 and P(B) = 0.5, with A and B independent. Find P(A and B) and P(A or B).
ChallengeA fair die is rolled 120 times. Find the expected number of sixes.
ChallengeA bag holds 3 red and 2 white balls. Two balls are drawn without replacement. Find the probability of at least one red ball.
Error redoError repair: a bag holds 4 red and 6 blue balls and two balls are drawn without replacement. A student answers four twenty-fifths for both being red. What is correct?

03　Answers and one-line notes

#	Answer	One-line note
1	1/4 (that is 0.25)	Five red out of twenty balls simplifies to one quarter.
2	1/2	Three even faces out of six simplify to one half.
3	1/13	Four aces in fifty-two cards simplify to one thirteenth.
4	1/10	Only seven among the numbers one to ten is a multiple of seven.
5	5/36	Five pairs make a total of eight out of thirty-six equally likely outcomes.
6	1/3	After one red ball is removed, three red remain among nine balls.
7	P(A and B) = 0.3; P(A or B) = 0.8	Multiply for both to get 0.3, then add and subtract the overlap for either.
8	20 times	Expected count is the number of trials times one sixth.
9	9/10	Both white is two fifths times one quarter, so the complement is nine tenths.
10	2/15	Four twenty-fifths treats the draws as independent; four tenths times three ninths simplifies to two fifteenths.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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