| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | E(X) and Var(X) with probability calculations |
| Difficulty | Moderate -0.3 This is a straightforward application of binomial distribution with clearly defined probabilities. Parts (i), (ii), and (iv) are direct textbook exercises requiring only formula substitution. Part (iii) requires recognizing the probability is (1/30) and applying the binomial formula, which is slightly less routine but still standard. The multi-part structure and clear setup make this slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| For correct final answer either as a decimal or a fraction | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| For using \((1/10)^k\) \(k>1\) | B1 | |
| For using a binomial expression with their 1/10 or seeing \(p^x*(1-p)^y\) | M1 | |
| For correct answer | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| For their \((1/10)^3\) seen | M1 | |
| For correct answer | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| For recognising \(n=5, p = 1/3\) | B1 | |
| For correct mean and variance, ft their n and p, \(p<1\) | B1 ft | 2 marks |
**(i)** $P(\text{no orange}) = (2/3)^3 = 0.132$ or $32/243$
| For correct final answer either as a decimal or a fraction | B1 | 1 mark
**(ii)** $P(\text{2 end in 6}) = (1/10)^2 \times (9/10)^3 \times {}_3C_2 = 0.0729$
| For using $(1/10)^k$ $k>1$ | B1
| For using a binomial expression with their 1/10 or seeing $p^x*(1-p)^y$ | M1
| For correct answer | A1 | 3 marks
**(iii)** $P(\text{2 orange end in 6}) = (1/30)^2 \times (29/30)^y \times {}_2C_2 = 0.0100$ accept 0.01
| For their $(1/10)^3$ seen | M1
| For correct answer | A1 | 2 marks
**(iv)** $n = 5, p = 1/3$; mean $= 5/3$, variance $= 10/9$
| For recognising $n=5, p = 1/3$ | B1
| For correct mean and variance, ft their n and p, $p<1$ | B1 ft | 2 marks5 A box contains 300 discs of different colours. There are 100 pink discs, 100 blue discs and 100 orange discs. The discs of each colour are numbered from 0 to 99 . Five discs are selected at random, one at a time, with replacement. Find\\
(i) the probability that no orange discs are selected,\\
(ii) the probability that exactly 2 discs with numbers ending in a 6 are selected,\\
(iii) the probability that exactly 2 orange discs with numbers ending in a 6 are selected,\\
(iv) the mean and variance of the number of pink discs selected.
\hfill \mbox{\textit{CAIE S1 2005 Q5 [8]}}