| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | P(a ≤ X ≤ b) range probability |
| Difficulty | Moderate -0.8 This is a straightforward application of the geometric distribution formula with minimal problem-solving required. Parts (i)-(iii) are direct substitutions into standard formulas, and part (iv) requires only basic enumeration of cases (X₁=1,X₂=2 or X₁=2,X₂=1) with independence. All techniques are routine for S1 level, making this easier than average. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| 0.8²×0.2 = \(\frac{16}{125}\) or 0.128 | M1, A1 2 | |
| 0.8²×0.2 + 0.8×0.2 + 0.8²×0.2 = \(\frac{976}{3125}\) or 0.312 (3 sfs) | M2, A1 3 | 1 term omitted or wrong or extra: M1. Using \(P(X<5)\) & \(P(X>2)\); three methods: 1 – \(0.8^3–(1–0.8^3)\) or 0.672 – 0.36: M2. Allow M1 for 1 – \(0.8^3–(1–0.8^3)\) or 0.672 – 0.488 or 1 – \(0.8^3–(1–0.8^3)\) or 0.5904 – 0.36. \(0.8^2 – 0.8^5\): M2 Allow M1 for \(0.8^3 – 0.8^5\) or \(0.8^2 – 0.8^4\). 0.2+0.8×0.2+0.8²×0.2+0.8³×0.2+0.8⁴×0.2 – (0.2+0.8×0.2): M2. One term omitted or wrong or extra: M1. But NB If include \(0.8^1×0.2\) in both \(P(X<5)\) & \(P(X<2)\), get correct ans but M1M0A0. M0 for eg 1 – \(0.8^5 – 0.8^2\) or 0.672 – 0.64 |
| 0.8²×0.2 = $\frac{16}{125}$ or 0.128 | M1, A1 2 | |
| 0.8²×0.2 + 0.8×0.2 + 0.8²×0.2 = $\frac{976}{3125}$ or 0.312 (3 sfs) | M2, A1 3 | 1 term omitted or wrong or extra: M1. Using $P(X<5)$ & $P(X>2)$; three methods: 1 – $0.8^3–(1–0.8^3)$ or 0.672 – 0.36: M2. Allow M1 for 1 – $0.8^3–(1–0.8^3)$ or 0.672 – 0.488 or 1 – $0.8^3–(1–0.8^3)$ or 0.5904 – 0.36. $0.8^2 – 0.8^5$: M2 Allow M1 for $0.8^3 – 0.8^5$ or $0.8^2 – 0.8^4$. 0.2+0.8×0.2+0.8²×0.2+0.8³×0.2+0.8⁴×0.2 – (0.2+0.8×0.2): M2. One term omitted or wrong or extra: M1. But NB If include $0.8^1×0.2$ in both $P(X<5)$ & $P(X<2)$, get correct ans but M1M0A0. M0 for eg 1 – $0.8^5 – 0.8^2$ or 0.672 – 0.64 |2 The random variable $X$ has the distribution $\operatorname { Geo } ( 0.2 )$. Find\\
(i) $\mathrm { P } ( X = 3 )$,\\
(ii) $\mathrm { P } ( 3 \leqslant X \leqslant 5 )$,\\
(iii) $\mathrm { P } ( X > 4 )$.
Two independent values of $X$ are found.\\
(iv) Find the probability that the total of these two values is 3 .
\hfill \mbox{\textit{OCR S1 2011 Q2 [11]}}