Worked solution
Combine a disc and a weighted die
An unbiased disc has a single dot marked on one side and two dots marked on the other side. A tetrahedral die has faces marked with score of 1, 2, 3, and 4. The probability of getting a score of 1, 2, 3, and 4 is and respectively, where .
A game is played by throwing the disc and the die together. The random variable is the sum of the score showing on the die and twice the number of dots showing on the disc.
(i) Find .
Given that ,
(ii) calculate the values of p and q, and
(iii) find the probability distribution of S.
[2017/PJC/II/6]
CJC · Discussion · Question 1
Answer
; , ; distribution in the table.
Construct the joint outcome table
The source solution models the disc and die as independent. This vertical form retains every cell of its two-by-four outcome table.
| Disc dots | Die score | S | Probability |
|---|---|---|---|
| 1 | 1 | 3 | |
| 1 | 2 | 4 | |
| 1 | 3 | 5 | |
| 1 | 4 | 6 | |
| 2 | 1 | 5 | |
| 2 | 2 | 6 | |
| 2 | 3 | 7 | |
| 2 | 4 | 8 |
(i) Two normalisation methods
Method 1 uses the die probabilities: , so . Method 2 sums all eight joint cells: , giving the same relation. Hence
(ii) Use the given probability
Only the one-dot/score-two cell gives S=4. Thus , so and .
(iii) Combine cells with equal S
| s | |
|---|---|
| 3 | |
| 4 | |
| 5 | |
| 6 | |
| 7 | |
| 8 |
The probabilities sum to one. S=5 and S=6 each collect two joint outcomes.
Common mistakes
Different joint outcomes can contribute to the same total score.
Solution notes
Full joint table, both normalisation methods and full final distribution retained. The independent-device assumption used by the source is explicit.