| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Single batch expected count |
| Difficulty | Moderate -0.8 This is a straightforward application of binomial distribution formulas with clear parameters given. Part (i) is direct calculation using B(3,0.87), part (ii) is simple multiplication of probability by 50, and part (iii) applies binomial distribution again with the probability from (i). All steps are routine with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure and nested probability in part (iii). |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=2) = \binom{3}{2} \times 0.87^2 \times 0.13 = 0.2952\) | M1 | \(0.87^2 \times 0.13\) |
| M1 | \(\binom{3}{2} \times p^2 q\) with \(p+q=1\) | |
| A1 CAO | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| In 50 throws expect \(50 \times (0.2952) = 14.76\) times | B1 FT | Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{two 20's twice}) = \binom{4}{2} \times 0.2952^2 \times 0.7048^2 = 0.2597\) | M1 | \(0.2952^2 \times 0.7048^2\) |
| A1 FT | their 0.2952. Total: 2 marks |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=2) = \binom{3}{2} \times 0.87^2 \times 0.13 = 0.2952$ | M1 | $0.87^2 \times 0.13$ |
| | M1 | $\binom{3}{2} \times p^2 q$ with $p+q=1$ |
| | A1 CAO | Total: 3 marks |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| In 50 throws expect $50 \times (0.2952) = 14.76$ times | B1 FT | Total: 1 mark |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{two 20's twice}) = \binom{4}{2} \times 0.2952^2 \times 0.7048^2 = 0.2597$ | M1 | $0.2952^2 \times 0.7048^2$ |
| | A1 FT | their 0.2952. Total: 2 marks |
---3 Douglas plays darts, and the probability that he hits the number he is aiming at is 0.87 for any particular dart.
Douglas aims a set of three darts at the number 20 ; the number of times he is successful can be modelled by $\mathrm { B } ( 3,0.87 )$.\\
(i) Calculate the probability that Douglas hits 20 twice.\\
(ii) Douglas aims fifty sets of 3 darts at the number 20. Find the expected number of sets for which Douglas hits 20 twice.\\
(iii) Douglas aims four sets of 3 darts at the number 20. Calculate the probability that he hits 20 twice for two sets out of the four.
\hfill \mbox{\textit{OCR MEI S1 Q3 [6]}}