| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Two-stage binomial problems |
| Difficulty | Standard +0.8 This is a two-stage probability problem requiring conditional probability and binomial distribution calculations. Parts (i) and (ii) are straightforward applications, but part (iii) requires summing over all four cases with different binomial parameters, demanding careful organization and computation. This goes beyond routine exercises and requires systematic problem-solving across multiple scenarios. |
| Spec | 2.03a Mutually exclusive and independent events2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(4, 2H) = \frac{1}{4} \times {}^4C_2 \times (\frac{1}{3})^2(\frac{2}{3})^2\) | M1 | Multiplying their 2H expression by \(\frac{1}{4}\) \([P(4)]\) |
| M1 | Remaining factor is \((\frac{1}{3})^2(\frac{2}{3})^2\) [or \(\frac{4}{81}\)] multiplied by integer value \(k \geq 1\) OE | |
| \(= \frac{2}{27}\) (0.0741) | A1 | |
| 3 total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(3, 3H) = \frac{1}{4} \times (\frac{1}{3})^3 = \frac{1}{108}\) (0.00926) | B1 | |
| 1 total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(1, 1H) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}\) (0.08333) | M1 | Correct expression for 1 of \(P(1, 1H)\), \(P(2, 2H)\), \(P(4, 4H)\); unsimplified (or better) |
| \(P(2, 2H) = \frac{1}{4} \times (\frac{1}{3})^2 = \frac{1}{36}\) (0.02778) | M1 | Summing their values for 3 or 4 appropriate outcomes for the 'game' with no additional outcomes |
| \(P(3, 3H) = \frac{1}{4} \times (\frac{1}{3})^3 = \frac{1}{108}\) (0.009259) | ||
| \(P(4, 4H) = \frac{1}{4} \times (\frac{1}{3})^4 = \frac{1}{324}\) (0.003086) | ||
| \(\text{Prob} = \frac{10}{81}\) (0.123) | A1 | |
| 3 total |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(4, 2H) = \frac{1}{4} \times {}^4C_2 \times (\frac{1}{3})^2(\frac{2}{3})^2$ | M1 | Multiplying their 2H expression by $\frac{1}{4}$ $[P(4)]$ |
| | M1 | Remaining factor is $(\frac{1}{3})^2(\frac{2}{3})^2$ [or $\frac{4}{81}$] multiplied by integer value $k \geq 1$ OE |
| $= \frac{2}{27}$ (0.0741) | A1 | |
| | **3 total** | |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(3, 3H) = \frac{1}{4} \times (\frac{1}{3})^3 = \frac{1}{108}$ (0.00926) | B1 | |
| | **1 total** | |
---
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(1, 1H) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$ (0.08333) | M1 | Correct expression for 1 of $P(1, 1H)$, $P(2, 2H)$, $P(4, 4H)$; unsimplified (or better) |
| $P(2, 2H) = \frac{1}{4} \times (\frac{1}{3})^2 = \frac{1}{36}$ (0.02778) | M1 | Summing their values for 3 or 4 appropriate outcomes for the 'game' with no additional outcomes |
| $P(3, 3H) = \frac{1}{4} \times (\frac{1}{3})^3 = \frac{1}{108}$ (0.009259) | | |
| $P(4, 4H) = \frac{1}{4} \times (\frac{1}{3})^4 = \frac{1}{324}$ (0.003086) | | |
| $\text{Prob} = \frac{10}{81}$ (0.123) | A1 | |
| | **3 total** | |
---4 A fair tetrahedral die has faces numbered $1,2,3,4$. A coin is biased so that the probability of showing a head when thrown is $\frac { 1 } { 3 }$. The die is thrown once and the number $n$ that it lands on is noted. The biased coin is then thrown $n$ times. So, for example, if the die lands on 3 , the coin is thrown 3 times.\\
(i) Find the probability that the die lands on 4 and the number of times the coin shows heads is 2 .\\
(ii) Find the probability that the die lands on 3 and the number of times the coin shows heads is 3 .\\
(iii) Find the probability that the number the die lands on is the same as the number of times the coin shows heads.\\
\hfill \mbox{\textit{CAIE S1 2017 Q4 [7]}}