| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | State advantages of diagram types |
| Difficulty | Easy -1.3 This is a straightforward two-way table completion followed by basic probability calculations. Part (i) requires simple arithmetic to fill in the table, parts (ii)-(iv) are direct probability readings from the table, and part (v) is a standard hypergeometric/combination probability calculation. All techniques are routine for S1 level with no problem-solving insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Wrapped in gold foil | Unwrapped | Total | |
| Chocolate-covered | |||
| Not chocolate-covered | |||
| Total | 30 |
| Answer | Marks | Guidance |
|---|---|---|
| wrapped | unwrapped | total |
| choc | 7 | 10 |
| not choc | 5 | 8 |
| total | 12 | 18 |
| B1 | One correct row or column numbers | |
| B1 [2] | All correct including labels | |
| (ii) 12/30 (0.4) | B1ft [1] | Ft their table |
| (iii) 10/18 (5/9) (0.556) | B1ft [1] | Ft their table |
| (iv) 10/17 (0.588) | B1ft [1] | Ft their table |
| (v) \(P(2 \text{ wrapped}) = \frac{12}{30} \times \frac{11}{29} \times \frac{18}{28} \times \frac{17}{27} \times {}_4C_2\) | M1 | Mult by \(_4C_2\) |
| M1 | 12 × 11 × 18 × 17 seen in num | |
| M1 | 30 × 29 × 28 × 27 seen in denom | |
| \(= 0.368\) (374/1015) | A1 | Correct answer |
| OR: \({}_{12}C_2 \times {}_{18}C_2 / {}_{30}C_4\) | M1 | \(_nC_2\) seen mult or alone in num (not added) |
| M1 | \(_nC_2\) seen mult or alone in num (not added) | |
| M1 | \(_nC_4\) seen in denom | |
| \(= 0.368\) | A1 [4] | Correct answer |
(i)
| | wrapped | unwrapped | total |
|---|---|---|---|
| choc | 7 | 10 | 17 |
| not choc | 5 | 8 | 13 |
| total | 12 | 18 | 30 |
| B1 | One correct row or column numbers
| B1 [2] | All correct including labels
(ii) 12/30 (0.4) | B1ft [1] | Ft their table
(iii) 10/18 (5/9) (0.556) | B1ft [1] | Ft their table
(iv) 10/17 (0.588) | B1ft [1] | Ft their table
(v) $P(2 \text{ wrapped}) = \frac{12}{30} \times \frac{11}{29} \times \frac{18}{28} \times \frac{17}{27} \times {}_4C_2$ | M1 | Mult by $_4C_2$
| M1 | 12 × 11 × 18 × 17 seen in num
| M1 | 30 × 29 × 28 × 27 seen in denom
$= 0.368$ (374/1015) | A1 | Correct answer
OR: ${}_{12}C_2 \times {}_{18}C_2 / {}_{30}C_4$ | M1 | $_nC_2$ seen mult or alone in num (not added)
| M1 | $_nC_2$ seen mult or alone in num (not added)
| M1 | $_nC_4$ seen in denom
$= 0.368$ | A1 [4] | Correct answer6 A box of biscuits contains 30 biscuits, some of which are wrapped in gold foil and some of which are unwrapped. Some of the biscuits are chocolate-covered. 12 biscuits are wrapped in gold foil, and of these biscuits, 7 are chocolate-covered. There are 17 chocolate-covered biscuits in total.\\
(i) Copy and complete the table below to show the number of biscuits in each category.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
& Wrapped in gold foil & Unwrapped & Total \\
\hline
Chocolate-covered & & & \\
\hline
Not chocolate-covered & & & \\
\hline
Total & & & 30 \\
\hline
\end{tabular}
\end{center}
A biscuit is selected at random from the box.\\
(ii) Find the probability that the biscuit is wrapped in gold foil.
The biscuit is returned to the box. An unwrapped biscuit is then selected at random from the box.\\
(iii) Find the probability that the biscuit is chocolate-covered.
The biscuit is returned to the box. A biscuit is then selected at random from the box.\\
(iv) Find the probability that the biscuit is unwrapped, given that it is chocolate-covered.
The biscuit is returned to the box. Nasir then takes 4 biscuits without replacement from the box.\\
(v) Find the probability that he takes exactly 2 wrapped biscuits.
\hfill \mbox{\textit{CAIE S1 2012 Q6 [9]}}