| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Game theory with alternating players |
| Difficulty | Standard +0.8 This question requires systematic enumeration of game sequences with conditional probability reasoning. Part (i) is straightforward, but parts (ii) and (iii) demand careful consideration of all possible winning sequences and application of conditional probability with multiple cases. The alternating wins structure and the 'given that' condition in part (iii) elevate this beyond routine geometric distribution problems, requiring problem-solving rather than formula application. |
| Spec | 2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{B champ}) = 0.7 \times 0.7 = 0.49\) | B1 | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{B champ}) = P(WW) + P(WLW) + P(LWW)\) | M1 | Summing at least 2 options, at least one of which is 3-factor |
| \(= (0.7{\times}0.7) + (0.7{\times}0.3{\times}0.7) + (0.3{\times}0.7{\times}0.7)\) | ||
| \(= 0.49 + 0.147 + 0.147\) | B1 | 0.147 seen, unsimplified |
| \(= 0.784\) | A1 | Correct answer — Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(T2\mid T) = \dfrac{P(T2 \cap T)}{P(T)}\) | M1 | Attempt \(P(T2 \cap T)\) seen anywhere, sum of 2 terms |
| \(= \dfrac{0.3 \times 0.3 + 0.7 \times 0.3 \times 0.3}{0.216}\) | A1 | Correct unsimplified numerator of a fraction |
| M1 | Dividing by their \((1 - \mathbf{(ii)}\ \checkmark)\) oe | |
| \(= 0.708\) | A1 | Correct answer — Total: 4 |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{B champ}) = 0.7 \times 0.7 = 0.49$ | B1 | **Total: 1** |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{B champ}) = P(WW) + P(WLW) + P(LWW)$ | M1 | Summing at least 2 options, at least one of which is 3-factor |
| $= (0.7{\times}0.7) + (0.7{\times}0.3{\times}0.7) + (0.3{\times}0.7{\times}0.7)$ | | |
| $= 0.49 + 0.147 + 0.147$ | B1 | 0.147 seen, unsimplified |
| $= 0.784$ | A1 | Correct answer — **Total: 3** |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(T2\mid T) = \dfrac{P(T2 \cap T)}{P(T)}$ | M1 | Attempt $P(T2 \cap T)$ seen anywhere, sum of 2 terms |
| $= \dfrac{0.3 \times 0.3 + 0.7 \times 0.3 \times 0.3}{0.216}$ | A1 | Correct unsimplified numerator of a fraction |
| | M1 | Dividing by their $(1 - \mathbf{(ii)}\ \checkmark)$ oe |
| $= 0.708$ | A1 | Correct answer — **Total: 4** |
---6 Tom and Ben play a game repeatedly. The probability that Tom wins any game is 0.3 . Each game is won by either Tom or Ben. Tom and Ben stop playing when one of them (to be called the champion) has won two games.\\
(i) Find the probability that Ben becomes the champion after playing exactly 2 games.\\
(ii) Find the probability that Ben becomes the champion.\\
(iii) Given that Tom becomes the champion, find the probability that he won the 2nd game.
\hfill \mbox{\textit{CAIE S1 2014 Q6 [8]}}