| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | At least one success |
| Difficulty | Moderate -0.3 Part (a) is a standard binomial 'at least one' question using the complement rule (1-P(none)), with part (ii) requiring simple trial-and-error to find n. Part (b) involves conditional probability without replacement but follows a straightforward pattern of alternating draws. All techniques are routine for S1 level with no novel insight required, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) \(P(\text{at least one 3}) = 1 - P(\text{no 3s}) = 1 - (5/6)^6 = 0.806\) | M1 | Using \(1 - \text{none}\) |
| A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(n > 12.6\) | B1 | |
| M1 | Equation or inequality involving \(n\) and 0.9 | |
| M1 | Solving attempt of sensible equation, can be trial | |
| \(n = 13\) | A1 [4] | Correct answer |
| (b) \(P(R \text{ wins his 1}^{\text{st}} \text{ ball}) = P(GY) = 15/56 (0.268)\) | M1 | Using \(P(GY)\) |
| \(P(R \text{ wins 2}^{\text{nd}} \text{ ball}) = P(GGGY) = 3/28\) | M1 | Attempt to find \(P(GGGY)\) or \(P(GGGGY)\) |
| \(P(R \text{ wins 3rd ball}) = P(GGGGGY)\) | M1 | Adding three options |
| \(\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times \frac{3}{3} = 1/56\) | ||
| \(P(R \text{ wins}) = 11/28 (0.393)\) | A1 [4] | Correct answer |
**(a)** **(i)** $P(\text{at least one 3}) = 1 - P(\text{no 3s}) = 1 - (5/6)^6 = 0.806$ | M1 | Using $1 - \text{none}$
| A1 [2] | Correct answer
**(ii)** $P(\text{at least 1 three}) = 1 - (5/6)^n$
$1 - (5/6)^n > 0.9$
$n > 12.6$ | B1 |
| M1 | Equation or inequality involving $n$ and 0.9
| M1 | Solving attempt of sensible equation, can be trial
$n = 13$ | A1 [4] | Correct answer
**(b)** $P(R \text{ wins his 1}^{\text{st}} \text{ ball}) = P(GY) = 15/56 (0.268)$ | M1 | Using $P(GY)$
$P(R \text{ wins 2}^{\text{nd}} \text{ ball}) = P(GGGY) = 3/28$ | M1 | Attempt to find $P(GGGY)$ or $P(GGGGY)$
$P(R \text{ wins 3rd ball}) = P(GGGGGY)$ | M1 | Adding three options
$\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times \frac{3}{3} = 1/56$ | |
$P(R \text{ wins}) = 11/28 (0.393)$ | A1 [4] | Correct answer7
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the probability of getting at least one 3 when 9 fair dice are thrown.
\item When $n$ fair dice are thrown, the probability of getting at least one 3 is greater than 0.9. Find the smallest possible value of $n$.
\end{enumerate}\item A bag contains 5 green balls and 3 yellow balls. Ronnie and Julie play a game in which they take turns to draw a ball from the bag at random without replacement. The winner of the game is the first person to draw a yellow ball. Julie draws the first ball. Find the probability that Ronnie wins the game.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2011 Q7 [10]}}