KCSE 2025 Maths P2 Q22 — Probability Space & Tree Diagram Without Replacement

KCSE 2025 Form 4 Probability

Published

The Question

“A bag contains five identical balls numbered 1 to 5. (a) Two balls are drawn at random, one at a time, without replacement. (i) Draw a probability space showing all possible pairs of numbers on the two balls drawn. (ii) Find the probability that both numbers drawn are greater than 3. (iii) Find the probability that the sum of the two numbers drawn does not exceed 6. (b) Three of the balls are green and the remaining two are red. For the two balls drawn in part (a), determine the probability that they were (i) of the same colour, and (ii) of mixed colours.”

1

Draw the probability space (a)(i)

List the first draw down the side and the second draw across the top, forming a five-by-five grid. Because the drawing is without replacement, you cannot pick the same ball twice, so cross out the whole diagonal where the two numbers are equal. Every remaining cell is one ordered pair, so counting them gives the size of the sample space.

5×5=25cells, minus the 5 on the diagonal5 \times 5 = 25 \quad \text{cells, minus the 5 on the diagonal}
n(S)=25−5=20 ordered outcomesn(S) = 25 - 5 = 20 \text{ ordered outcomes}

Order matters here because the balls are drawn one after another, so (4, 5) and (5, 4) are two different outcomes.

2

Both numbers greater than 3 (a)(ii)

Only the balls numbered 4 and 5 are greater than 3, so both draws must come from just those two balls. The ordered outcomes that work are 4 then 5 and 5 then 4. Divide the number of favourable outcomes by the 20 in the sample space and simplify.

Favourable: (4,5), (5,4)⇒2 outcomes\text{Favourable: } (4,5),\ (5,4) \Rightarrow 2 \text{ outcomes}
P(both>3)=220=110P(\text{both} > 3) = \frac{2}{20} = \frac{1}{10}
3

Sum does not exceed 6 (a)(iii)

You need the two distinct numbers to add up to 6 or less. The unordered pairs that qualify are (1,2), (1,3), (1,4), (1,5), (2,3) and (2,4). Each of these can occur in two orders on the grid, so there are twelve favourable ordered outcomes. Divide by 20 and simplify.

{1,2},{1,3},{1,4},{1,5},{2,3},{2,4}⇒6 pairs\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{2,3\},\{2,4\} \Rightarrow 6 \text{ pairs}
6×2=12 ordered outcomes6 \times 2 = 12 \text{ ordered outcomes}
P(sum≤6)=1220=35P(\text{sum} \le 6) = \frac{12}{20} = \frac{3}{5}
4

Build the probability tree for colour (b)

Now use the colours: three green and two red. On the first draw green has probability 3/5 and red 2/5. After a green is removed, four balls remain with two green and two red; after a red is removed, four remain with three green and one red. Every second-draw fraction therefore has 4 on the bottom, which keeps the arithmetic tidy.

P(G)=35,P(R)=25P(G) = \tfrac{3}{5}, \quad P(R) = \tfrac{2}{5}
After G: P(G)=24, P(R)=24\text{After G: } P(G) = \tfrac{2}{4},\ P(R) = \tfrac{2}{4}
After R: P(G)=34, P(R)=14\text{After R: } P(G) = \tfrac{3}{4},\ P(R) = \tfrac{1}{4}
5

Same colour (b)(i)

Same colour means green then green, or red then red. Multiply the probabilities along each branch of the tree, then add the two branch results together because either one satisfies the event.

P(GG)=35×24=620=310P(GG) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}
P(RR)=25×14=220=110P(RR) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = \frac{1}{10}
P(same)=310+110=410=25P(\text{same}) = \frac{3}{10} + \frac{1}{10} = \frac{4}{10} = \frac{2}{5}
6

Mixed colours (b)(ii)

Mixed colours is the exact opposite of same colour: the two balls are either both green or both red, or they are mixed, with no other possibility. So subtract the same-colour probability from 1 rather than adding up the mixed branches separately.

P(mixed)=1−P(same)=1−25=35P(\text{mixed}) = 1 - P(\text{same}) = 1 - \frac{2}{5} = \frac{3}{5}

Final Result

There are 20 ordered outcomes in the probability space. The probability both numbers exceed 3 is 1/10. The probability the sum does not exceed 6 is 3/5. For colour, the probability of same colour is 2/5 and the probability of mixed colours is 3/5.

Why this method works

Drawing without replacement shrinks the pool for the second pick, which is why the diagonal is removed from the grid and why the tree's second-draw fractions have a denominator of 4. Because the outcomes on the grid are all equally likely, each probability is simply favourable outcomes over 20. On the tree, multiplying along a branch gives the chance of that ordered sequence, and adding branches combines mutually exclusive ways an event can happen. Same colour and mixed colour together cover every possibility, so their probabilities must sum to 1, which lets you get the mixed case by subtraction.

✓

The colour probabilities add up: same colour 2/5 plus mixed 3/5 equals 5/5 = 1, confirming no outcome is missed or double-counted.