| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Standard +0.3 This is a straightforward probability question requiring systematic enumeration of outcomes (5×5=25 cases), construction of a probability distribution table, and calculation of mean. Parts (iv)-(v) involve simple geometric probability with one play resulting in a win/draw. All techniques are standard S1 material with no novel insight required, making it slightly easier than average. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\text{P}(6) = \text{P}(3, 9) + \text{P}(9, 3) = 2/25 = 0.08\) AG | B1 [1] | Accept \(2/25\) seen |
| (ii) | M1, A1 [2] | Values \(0 - 6\) seen could be in list; All correct |
| x | 0 | 1 |
| Prob | 0.2 | 0.24 |
| (iii) Mean \(= \Sigma xp = 2.56\) \((64/25)\) | B1 [1] | |
| (iv) \(\text{P}(4, 5, 6) = 0.4(10/25)\) or \(0.16 + 0.16 + 0.08\) | B1 ft M1 | If their P(4, 5, 6) providing \(p < 1\); Multiplying by their P(draw) providing \(p < 1\) |
| \(= \text{P(draw)} \times 0.4\) \(= 0.2 \times 0.4 = 0.08\) \((2/25)\) | A1 ft [3] | Correct answer |
| (v) \(\text{P(J wins on }nth\text{ go)} = (0.2)^{r-1} \times 0.4\) oe | M1, A1 ft [2] | Mult by any \(p^r\) or \(p^{r-1}\), \(p < 1\); fi their probs |
**(i)** $\text{P}(6) = \text{P}(3, 9) + \text{P}(9, 3) = 2/25 = 0.08$ AG | B1 [1] | Accept $2/25$ seen
**(ii)** | M1, A1 [2] | Values $0 - 6$ seen could be in list; All correct
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Prob | 0.2 | 0.24 | 0.08 | 0.08 | 0.16 | 0.16 | 0.08 |
**(iii)** Mean $= \Sigma xp = 2.56$ $(64/25)$ | B1 [1] |
**(iv)** $\text{P}(4, 5, 6) = 0.4(10/25)$ or $0.16 + 0.16 + 0.08$ | B1 ft M1 | If their P(4, 5, 6) providing $p < 1$; Multiplying by their P(draw) providing $p < 1$
$= \text{P(draw)} \times 0.4$ $= 0.2 \times 0.4 = 0.08$ $(2/25)$ | A1 ft [3] | Correct answer
**(v)** $\text{P(J wins on }nth\text{ go)} = (0.2)^{r-1} \times 0.4$ oe | M1, A1 ft [2] | Mult by any $p^r$ or $p^{r-1}$, $p < 1$; fi their probs7 Judy and Steve play a game using five cards numbered 3, 4, 5, 8, 9. Judy chooses a card at random, looks at the number on it and replaces the card. Then Steve chooses a card at random, looks at the number on it and replaces the card. If their two numbers are equal the score is 0 . Otherwise, the smaller number is subtracted from the larger number to give the score.\\
(i) Show that the probability that the score is 6 is 0.08 .\\
(ii) Draw up a probability distribution table for the score.\\
(iii) Calculate the mean score.
If the score is 0 they play again. If the score is 4 or more Judy wins. Otherwise Steve wins. They continue playing until one of the players wins.\\
(iv) Find the probability that Judy wins with the second choice of cards.\\
(v) Find an expression for the probability that Judy wins with the $n$th choice of cards.
\hfill \mbox{\textit{CAIE S1 2011 Q7 [9]}}