| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Finding binomial parameters from properties |
| Difficulty | Moderate -0.8 This is a straightforward application of basic probability and binomial distribution. Part (i) requires simple algebraic manipulation (if P(HH)=p²=0.04, then P(TT)=(1-p)²). Part (ii) involves solving a quadratic equation from 2p(1-p)=0.42. Both parts are routine exercises with clear methods and minimal steps, making this easier than average for A-level. |
| Spec | 2.03a Mutually exclusive and independent events |
| Answer | Marks |
|---|---|
| \((1 - \text{their } \sqrt{0.04})^2 = 0.64\) | M1 |
| M1 | |
| A1 | .3. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2±\sqrt{(2)^2 - 4×0.42}}{2}\) or \(\frac{1±\sqrt{(1)^2 - 4×0.21}}{2 × 1}\) | M1 | \(2pq = 0.42\) or \(pq = 0.21\) Allow \(pq=0.42\) or opp signs, correct terms any order (= 0) |
| M1 | oe Correct | |
| Dep B1M1 Any corr substn or fact'n |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 0.7\) or \(0.3\) | M1 | Al 5 |
| B1 | ||
| M1 |
| Answer | Marks |
|---|---|
| M1 |
| Answer | Marks |
|---|---|
| \(p^2 - p + 0.25 = -0.21 + 0.25\) oe | M1 |
| Answer | Marks |
|---|---|
| \(p = 0.3\) or \(0.7\) | A1 |
| Answer | Marks |
|---|---|
| M1 |
| Answer | Marks |
|---|---|
| M1 |
| Answer | Marks |
|---|---|
| \(p^2 - p + 0.25 = -0.21 + 0.25\) oe | M1 |
| Answer | Marks |
|---|---|
| \(p = 0.3\) or \(0.7\) | A1 |
| M1 |
| Answer | Marks |
|---|---|
| \(p^2 + q^2 = 0.58\) | B1 |
| Answer | Marks |
|---|---|
| corr substn or fact'n | M1 |
| M1 | |
| ( \(= \frac{p}{2p - p^2}\) ) \(= \frac{p}{p(2-p)}\) | |
| ( \(= \frac{1}{2 - p}\) ) \(= \frac{1}{2-(1-q)}\) |
### Part i
$\sqrt{0.04}$ (= 0.2)
$(1 - \text{their } \sqrt{0.04})^2 = 0.64$ | M1 |
| M1 |
| A1 | .3.
### Part ii
$1 - p$ seen
$2p(1 - p) = 0.42$ or $p(1 - p) = 0.21$ oe
$2p^2 - 2p + 0.42(= 0)$ or $p^2 - p - 0.21(= 0)$
$\frac{2±\sqrt{(2)^2 - 4×0.42}}{2}$ or $\frac{1±\sqrt{(1)^2 - 4×0.21}}{2 × 1}$ | M1 | $2pq = 0.42$ or $pq = 0.21$ Allow $pq=0.42$ or opp signs, correct terms any order (= 0)
| M1 | oe Correct
| | Dep B1M1 Any corr substn or fact'n
or $(p - 0.7)(p - 0.3) = 0$ or $(10p - 7)(10p - 3) = 0$
$p = 0.7$ or $0.3$ | M1 | Al 5 |
| B1 |
| M1 |
$2p(1 - p) = 0.42$ or $p(1 - p) = 0.21$ M1
| M1 |
$p^2 - p = -0.21$
$p^2 - p + 0.25 = -0.21 + 0.25$ oe | M1 |
$(p - 0.5)^2 = 0.04$
$(p - 0.5) = ±0.02$
$p = 0.3$ or $0.7$ | A1 |
---
| M1 |
$1 - p$ seen
$2p(1 - p) = 0.42$ or $p(1 - p) = 0.21$ M1
| M1 |
$p^2 - p = -0.21$
$p^2 - p + 0.25 = -0.21 + 0.25$ oe | M1 |
$(p - 0.5)^2 = 0.04$
$(p - 0.5) = ±0.02$
$p = 0.3$ or $0.7$ | A1 |
| M1 |
$p^2 + 2pq + q^2 = 1$
$p^2 + q^2 = 0.58$ | B1 |
$p = 0.21/q$
$p^2 - 0.58p^2 + 0.0441 = 0$
corr substn or fact'n | M1 |
| M1 |
| | ( $= \frac{p}{2p - p^2}$ ) $= \frac{p}{p(2-p)}$
| | ( $= \frac{1}{2 - p}$ ) $= \frac{1}{2-(1-q)}$
**Total: 8**
---8 (i) A biased coin is thrown twice. The probability that it shows heads both times is 0.04 . Find the probability that it shows tails both times.\\
(ii) A nother coin is biased so that the probability that it shows heads on any throw is p . The probability that the coin shows heads exactly once in two throws is 0.42 . Find the two possible values of p.
\hfill \mbox{\textit{OCR S1 2007 Q8 [8]}}