| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard Bayes with discrete events |
| Difficulty | Moderate -0.8 This is a straightforward application of the law of total probability and Bayes' theorem with clearly stated probabilities and standard binomial distribution calculations. All parts follow textbook methods with no novel insight required—easier than average for A-level statistics. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(H) = P(BH) + P(SH) = 0.6 \times 0.05 + 0.4 \times 0.75\) | M1 | Summing two 2-factor probs using 0.6 with 0.05 or 0.95, and 0.4 with 0.75 or 0.25 |
| \(= 0.330\) or \(\frac{33}{100}\) | A1 | Correct final answer, accept 0.33 |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(S | H) = \frac{P(S \cap H)}{P(H)} = \frac{0.4 \times 0.75}{0.33} = \frac{0.3}{0.33}\) | M1 FT |
| \(= \frac{10}{11}\) or \(0.909\) | A1 | |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Var}(B) = 45 \times 0.6 \times 0.4\); \(\text{Var}(S) = 45 \times 0.4 \times 0.6\) | B1 | One variance stated unsimplified |
| Variances same | B1 | Second variance stated unsimplified and at least one variance clearly identified, and both evaluated *or* showing equal *or* conclusion made. SR B1 – Standard Deviation calculated fulfilling all criteria for variance method but calculated to SD |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - P(0,1) = 1 - [(0.6)^{10} + {}^{10}C_1(0.4)(0.6)^9]\) | M1 | Bin term \({}^{10}C_x p^x (1-p)^{10-x}\), \(0 < p < 1\) |
| OR \(P(2,3,4,5,6,7,8,9,10) = {}^{10}C_2(0.4)^2(0.6)^8 + \ldots + {}^{10}C_9(0.4)^9(0.6) + (0.4)^{10}\) | M1 | Correct unsimplified answer |
| \(= 0.954\) | A1 | |
| Total: | 3 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(H) = P(BH) + P(SH) = 0.6 \times 0.05 + 0.4 \times 0.75$ | M1 | Summing two 2-factor probs using 0.6 with 0.05 or 0.95, and 0.4 with 0.75 or 0.25 |
| $= 0.330$ or $\frac{33}{100}$ | A1 | Correct final answer, accept 0.33 |
| **Total:** | **2** | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(S|H) = \frac{P(S \cap H)}{P(H)} = \frac{0.4 \times 0.75}{0.33} = \frac{0.3}{0.33}$ | M1 FT | Their $\frac{P(S \cap H)}{P(H)}$ unsimplified, FT from (i) |
| $= \frac{10}{11}$ or $0.909$ | A1 | |
| **Total:** | **2** | |
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Var}(B) = 45 \times 0.6 \times 0.4$; $\text{Var}(S) = 45 \times 0.4 \times 0.6$ | B1 | One variance stated unsimplified |
| Variances same | B1 | Second variance stated unsimplified **and** at least one variance clearly identified, **and** both evaluated *or* showing equal *or* conclusion made. SR B1 – Standard Deviation calculated fulfilling all criteria for variance method but calculated to SD |
| **Total:** | **2** | |
## Question 7(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - P(0,1) = 1 - [(0.6)^{10} + {}^{10}C_1(0.4)(0.6)^9]$ | M1 | Bin term ${}^{10}C_x p^x (1-p)^{10-x}$, $0 < p < 1$ |
| OR $P(2,3,4,5,6,7,8,9,10) = {}^{10}C_2(0.4)^2(0.6)^8 + \ldots + {}^{10}C_9(0.4)^9(0.6) + (0.4)^{10}$ | M1 | Correct unsimplified answer |
| $= 0.954$ | A1 | |
| **Total:** | **3** | |7 During the school holidays, each day Khalid either rides on his bicycle with probability 0.6 , or on his skateboard with probability 0.4 . Khalid does not ride on both on the same day. If he rides on his bicycle then the probability that he hurts himself is 0.05 . If he rides on his skateboard the probability that he hurts himself is 0.75 .\\
(i) Find the probability that Khalid hurts himself on any particular day.\\
(ii) Given that Khalid hurts himself on a particular day, find the probability that he is riding on his skateboard.\\
(iii) There are 45 days of school holidays. Show that the variance of the number of days Khalid rides on his skateboard is the same as the variance of the number of days that Khalid rides on his bicycle.\\
(iv) Find the probability that Khalid rides on his skateboard on at least 2 of 10 randomly chosen days in the school holidays.\\
\hfill \mbox{\textit{CAIE S1 2017 Q7 [9]}}