KCSE Mathematics Probability: Worked Examples With and Without Replacement

In KCSE Mathematics probability questions, one short instruction can change the entire calculation: the object is replaced, or the object is not replaced. Before multiplying fractions, decide what remains available for the next selection. These original worked examples show how to record each possibility, choose the correct paths and check your answer.

You will practise drawing with and without replacement, the meaning of AND and OR, independent events and the complement method. All questions and solutions are on this page. The tables give the complete branches of two probability trees in text, so you can follow them without an image or a download.

Start with the experiment and the event

An experiment is the random process being carried out, such as selecting two counters. An outcome is one possible result. An event is the result or collection of results you want to count. “Green then yellow” is one ordered colour outcome; “one counter of each colour” includes two ordered outcomes.

When the individual objects are equally likely to be selected, the probability of an event is the number of favourable objects divided by the total number available. Five red counters and three blue counters do not make red and blue equally likely just because there are two colours. The counts matter.

For every question, identify these details before calculating:

• What is being selected, tossed or spun?
• How many selections or trials take place?
• Is an object returned before the next selection?
• Does the question specify an order, exactly one result or at least one result?
• What makes each individual outcome equally likely?

Use fractions during your working and simplify at the end. Repeatedly rounding decimals can turn an exact result into an inaccurate one.

Worked example 1: drawing with replacement

A bag contains 7 green counters and 3 yellow counters. The counters are identical apart from colour. A counter is selected at random, its colour is recorded, and it is returned to the bag. The bag is mixed before a second random selection.

Let G mean green and Y mean yellow. On the first draw, P(G) = 7/10 and P(Y) = 3/10. Replacement restores the bag to its original contents. Therefore the second-draw probabilities are again 7/10 and 3/10, whichever colour appeared first.

Read each row of the following tree table from left to right. The first two columns describe the two successive branches; the final column gives the probability of following that complete path.

Find the probability that both counters are green

Only GG satisfies this condition. Multiply along that path: P(both green) = 7/10 × 7/10 = 49/100. Adding 7/10 and 7/10 would give more than 1 and would not describe two successive green selections.

Find the probability of exactly one yellow counter

The yellow counter may appear first or second. Include GY and YG, and exclude YY because that contains two yellow counters. The two complete paths cannot occur together in the same two-draw experiment, so add them:

P(exactly one yellow) = 21/100 + 21/100 = 42/100 = 21/50.

Check the complete tree

Every possible colour sequence is included once. The path probabilities total (49 + 21 + 21 + 9)/100 = 1. At each branching point, the outgoing probabilities also total 7/10 + 3/10 = 1. These are two different checks, and both should work.

Worked example 2: drawing without replacement

Use the same initial bag of 7 green and 3 yellow counters. This time, the first counter is kept outside the bag before the second is selected. There are now only 9 counters on the second draw, but the remaining colour counts depend on the first result.

• After a green first draw, 6 green and 3 yellow counters remain.
• After a yellow first draw, 7 green and 2 yellow counters remain.

This is why reducing the denominator alone is insufficient. The numerator must match the colour count on that particular branch.

The probability of both green is now 7/15. This is smaller than 49/100 in the replacement example: removing a green counter makes a second green less likely. That comparison is a useful reasonableness check.

For exactly one yellow, add GY and YG: 7/30 + 7/30 = 7/15. For two counters of the same colour, add GG and YY: 7/15 + 1/15 = 8/15.

The four paths total (42 + 21 + 21 + 6)/90 = 1. After a green first draw, 6/9 + 3/9 = 1; after a yellow first draw, 7/9 + 2/9 = 1. Notice that each pair of second branches has its own correct counts.

AND, OR and the overlap warning

For successive events, multiply along a complete path. The second probability must describe the situation after the first event. That is why 7/10 × 6/9 is correct for GG without replacement. Multiplication is not limited to independent events; dependent events require the appropriate conditional probability.

For alternative complete paths that cannot happen together, add their probabilities. GY and YG are separate alternatives. Their probabilities can be added directly.

However, the word OR does not always mean “add without adjustment”. If events overlap, direct addition counts the shared outcomes twice. The general rule is P(A or B) = P(A) + P(B) − P(A and B).

For example, roll a fair six-sided die once. Let A be “an even number” and B be “a number greater than 4”. A contains 2, 4 and 6; B contains 5 and 6. The number 6 belongs to both. Thus P(A or B) = 3/6 + 2/6 − 1/6 = 2/3. Listing the distinct favourable results, 2, 4, 5 and 6, gives the same answer.

Use the complement for “at least one”

The complement of an event contains all outcomes where that event does not happen. Since the event and its complement cover the whole experiment, their probabilities add to 1.

In two draws, “at least one green” means one green or two greens. Its complement is no green, which means YY in our two-colour bag. “Exactly one green” is not its complement.

• With replacement: P(at least one green) = 1 − 9/100 = 91/100.
• Without replacement: P(at least one green) = 1 − 1/15 = 14/15.

You could add GG, GY and YG instead. The complement route is shorter because it calculates only one excluded path. With three or more selections, that saving can be even greater.

Worked example 3: two independent devices

A fair spinner has four equal sectors: three purple and one orange. A fair six-sided die is rolled independently of the spin. Find the probability of purple and a number greater than 4.

P(purple) = 3/4. The favourable die results are 5 and 6, so P(greater than 4) = 2/6. Independence means that learning the spinner result does not change the die probabilities.

P(purple and greater than 4) = 3/4 × 2/6 = 6/24 = 1/4. The assumptions about fairness, equal sectors and independence are part of the question. Without them, counting sector labels alone would not justify this calculation.

Independent practice: attempt before reading the solutions

1. A bag has 4 red and 5 blue counters. Two random draws are made with replacement and mixing. Find the probability of both red, exactly one red, and at least one red.
2. A bag has 6 white and 2 black counters. Two counters are drawn at random without replacement. Find the probability of both white, exactly one black, and at least one black.
3. A fair coin is tossed and a fair six-sided die is rolled independently. Find the probability of a head and a multiple of 3.
4. A box has 5 silver, 3 bronze and 2 gold tokens, identical apart from colour. Two random draws are made without replacement. Find the probability of both silver and of at least one gold.

Solution 1: replacement and three different events

Both red: 4/9 × 4/9 = 16/81. Exactly one red: red then blue or blue then red, so (4/9 × 5/9) + (5/9 × 4/9) = 40/81. At least one red: exclude two blues, giving 1 − (5/9 × 5/9) = 56/81. Check that both-red plus exactly-one-red is 16/81 + 40/81 = 56/81.

Solution 2: keep the second-draw counts consistent

Both white: 6/8 × 5/7 = 30/56 = 15/28. Exactly one black: (6/8 × 2/7) + (2/8 × 6/7) = 24/56 = 3/7. At least one black: exclude both white, giving 1 − 15/28 = 13/28. The difference between the last two answers is the probability of two black counters, 2/8 × 1/7 = 1/28.

Solution 3: count the favourable die faces

The multiples of 3 are 3 and 6. Therefore P(head and multiple of 3) = 1/2 × 2/6 = 1/6. Do not use 3/6 merely because the event mentions the number 3.

Solution 4: combine colours for the complement

Both silver: 5/10 × 4/9 = 20/90 = 2/9. For at least one gold, group silver and bronze as the 8 non-gold tokens. P(no gold) = 8/10 × 7/9 = 28/45. Therefore P(at least one gold) = 1 − 28/45 = 17/45. Grouping non-gold tokens avoids listing every silver–bronze arrangement.

A final probability answer-check routine

1. Read the replacement instruction again. Check the second-draw denominator and each remaining colour count.
2. Name the event you calculated. “First green, second yellow” is narrower than “one of each colour”.
3. Check bounds. Every probability must lie from 0 to 1 inclusive.
4. Check completeness. All full tree paths should total 1, and outgoing branches at each point should total 1.
5. Use a second method when practical. Compare a complement answer with the sum of its favourable paths.

After correcting these questions, choose another exercise from KCSE Mathematics topic tests revision. That existing resource page requires a member login; it is not a free-download link. The worked examples and answers above can be used without leaving this lesson.

Frequently asked questions

Further reading

For the terminology behind these original examples, see OpenStax on independent and mutually exclusive events and its probability rules overview.