| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Mean/expectation of geometric distribution |
| Difficulty | Moderate -0.8 This is a straightforward application of the geometric distribution with clearly stated parameters (p=1/5). Parts (i)-(iii) require only direct recall of the distribution's definition, basic probability calculation P(X=3), and complementary probability P(X>4). No problem-solving insight or multi-step reasoning is needed—just routine application of standard formulas. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Geometric distribution | B1 | For 'geometric' or 'Geo(...)' stated |
| \(p = \frac{1}{5}\) | B1 | For correct parameter value |
| Each packet is equally likely to contain any of the 5 animals, independently of other packets | B1 | For either 'equally likely' or 'independent' |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{4}{5}\right)^4 \times \left(\frac{1}{5}\right) = \frac{256}{3125}\) or 0.128 | M1 | For any numerical '\(q^r p\)' calculation |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{4}{5}\right)^4\) or \(1 - \left\{\frac{1}{5} + \left(\frac{4}{5}\right)\left(\frac{1}{5}\right) + \left(\frac{4}{5}\right)^2\left(\frac{1}{5}\right) + \left(\frac{4}{5}\right)^3\left(\frac{1}{5}\right)\right\}\) | M1 | Allow M mark even if there is an error of 1 in the number of terms |
| \(\frac{256}{625}\) or 0.4096 or 0.410 | A1 | For correct expression for the answer |
| A1 | For correct answer |
**Part (i)**
Geometric distribution | B1 | For 'geometric' or 'Geo(...)' stated
$p = \frac{1}{5}$ | B1 | For correct parameter value
Each packet is equally likely to contain any of the 5 animals, independently of other packets | B1 | For either 'equally likely' or 'independent'
**Part (ii)**
$\left(\frac{4}{5}\right)^4 \times \left(\frac{1}{5}\right) = \frac{256}{3125}$ or 0.128 | M1 | For any numerical '$q^r p$' calculation
| A1 | For correct answer
**Part (iii)**
$\left(\frac{4}{5}\right)^4$ or $1 - \left\{\frac{1}{5} + \left(\frac{4}{5}\right)\left(\frac{1}{5}\right) + \left(\frac{4}{5}\right)^2\left(\frac{1}{5}\right) + \left(\frac{4}{5}\right)^3\left(\frac{1}{5}\right)\right\}$ | M1 | Allow M mark even if there is an error of 1 in the number of terms
$\frac{256}{625}$ or 0.4096 or 0.410 | A1 | For correct expression for the answer
| A1 | For correct answer
---4 Each packet of the breakfast cereal Fizz contains one plastic toy animal. There are five different animals in the set, and the cereal manufacturers use equal numbers of each. Without opening a packet it is impossible to tell which animal it contains. A family has already collected four different animals at the start of a year and they now need to collect an elephant to complete their set. The family is interested in how many packets they will need to buy before they complete their set.\\
(i) Name an appropriate distribution with which to model this situation. State the value(s) of any parameter(s) of the distribution, and state also any assumption(s) needed for the distribution to be a valid model.\\
(ii) Find the probability that the family will complete their set with the third packet they buy after the start of the year.\\
(iii) Find the probability that, in order to complete their collection, the family will need to buy more than 4 packets after the start of the year.
\hfill \mbox{\textit{OCR S1 Q4 [8]}}