| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Combined probability with other distributions |
| Difficulty | Moderate -0.8 Part (a) is a straightforward hypergeometric distribution requiring calculation of three probabilities using combinations. Part (b) is a direct binomial probability calculation using P(X ≤ 7) = 1 - P(X = 8) - P(X = 9) - P(X = 10). Both parts involve standard probability distribution techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Probability distribution table with correct outcome values (\(x = 0,1,2,3\) with \(P = \frac{1}{56}, \frac{15}{56}, \frac{30}{56}, \frac{10}{56}\)) | B1 | |
| \(P(0) = \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{1}{56}\), \(P(1) = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times 3 = \frac{15}{56}\), \(P(2) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times 3 = \frac{30}{56}\), \(P(3) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{10}{56}\) | M1 | M1 for denominator \(8\times7\times6\) |
| Any one probability correct (with correct outcome) | A1 | |
| All probabilities correct | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - P(8,9,10) = 1 - \left[{}^{10}C_8\, 0.64^8 0.36^2 + {}^{10}C_9\, 0.64^9 0.36^1 + 0.64^{10}\right]\) | M1 | |
| \(1 - (0.164156 + 0.064852 + 0.11529)\) | M1 | |
| \(0.759\) | A1 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Probability distribution table with correct outcome values ($x = 0,1,2,3$ with $P = \frac{1}{56}, \frac{15}{56}, \frac{30}{56}, \frac{10}{56}$) | B1 | |
| $P(0) = \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{1}{56}$, $P(1) = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times 3 = \frac{15}{56}$, $P(2) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times 3 = \frac{30}{56}$, $P(3) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{10}{56}$ | M1 | M1 for denominator $8\times7\times6$ |
| Any one probability correct (with correct outcome) | A1 | |
| All probabilities correct | A1 | |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - P(8,9,10) = 1 - \left[{}^{10}C_8\, 0.64^8 0.36^2 + {}^{10}C_9\, 0.64^9 0.36^1 + 0.64^{10}\right]$ | M1 | |
| $1 - (0.164156 + 0.064852 + 0.11529)$ | M1 | |
| $0.759$ | A1 | |3 A company produces small boxes of sweets that contain 5 jellies and 3 chocolates. Jemeel chooses 3 sweets at random from a box.
\begin{enumerate}[label=(\alph*)]
\item Draw up the probability distribution table for the number of jellies that Jemeel chooses.\\
The company also produces large boxes of sweets. For any large box, the probability that it contains more jellies than chocolates is 0.64 . 10 large boxes are chosen at random.
\item Find the probability that no more than 7 of these boxes contain more jellies than chocolates.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q3 [7]}}