KJSEA 2025 Maths Q40 — Probability Space (Coin and Die)

KJSEA 2025 Grade 9 Statistics

Published

The Question

“Rebecca tossed a coin and a die at the same time. (a) Write down the probability space showing all the possible outcomes. (b) Find the probability that she gets a head on the coin and a four on the die.”

1

List the outcomes of each object

Start by writing down what each object can show on its own. A coin has two faces, so it can land showing a head or a tail. A fair die has six faces, so it can show any whole number from one to six. These are the building blocks for every combined outcome.

Coin: {H,T}\text{Coin: } \{H, T\}
Die: {1,2,3,4,5,6}\text{Die: } \{1, 2, 3, 4, 5, 6\}
2

Pair every coin result with every die result

Because the coin and die are tossed at the same time, each single outcome is a pair: one coin result together with one die result. Match each of the two coin results with each of the six die results. This gives the full probability space, also called the sample space.

{H1,H2,H3,H4,H5,H6, T1,T2,T3,T4,T5,T6}\{H1, H2, H3, H4, H5, H6,\, T1, T2, T3, T4, T5, T6\}

Two coin results times six die results gives 2 x 6 = 12 outcomes.

3

Confirm the outcomes are equally likely

The coin is fair and the die is fair, so no single pairing is favoured over another. All twelve outcomes have exactly the same chance of happening. This matters because it lets us find a probability simply by counting.

Total equally likely outcomes=12\text{Total equally likely outcomes} = 12
4

Pick out the outcome we want and count it

We want a head on the coin and a four on the die at the same time. Look through the list for that exact pairing. Only one outcome matches, the pair head-and-four. So the number of favourable outcomes is one.

Favourable outcome: {H4}\text{Favourable outcome: } \{H4\}
5

Write the probability as favourable over total

For equally likely outcomes, the probability is the number of favourable outcomes divided by the total number of outcomes. With one favourable outcome out of twelve, the probability is one twelfth.

P(head and four)=favourabletotal=112P(\text{head and four}) = \frac{\text{favourable}}{\text{total}} = \frac{1}{12}

Final Result

(a) The probability space is {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} — twelve equally likely outcomes. (b) The probability of a head and a four is 1/12.

Why this method works

Tossing the coin and the die together are independent actions — the coin result does not affect the die and vice versa. When two independent choices are combined, the total number of outcomes is the product of the separate numbers of outcomes, which is why two coin faces and six die faces produce twelve combined pairs. Because every pairing is equally likely, probability reduces to simple counting: divide the number of pairings you want by the total number of pairings. Only the single pair head-and-four fits the requirement, so the probability is one out of twelve.

✓

P(head) = 1/2 and P(four) = 1/6, and for independent events these multiply: 1/2 x 1/6 = 1/12, which matches the counting answer.