| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Multiple independent trials |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability question involving independent events and basic probability rules. Part (a) requires simple multiplication of independent probabilities and addition of mutually exclusive cases. Part (b) involves conditional probability but with clearly stated probabilities requiring only multiplication and addition. All calculations are routine with no conceptual challenges or novel problem-solving required. |
| Spec | 2.03a Mutually exclusive and independent events |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Walk | Cycle | Bus |
| Rita | 0.65 | 0.10 | 0.25 |
| Said | 0.40 | 0.45 | 0.15 |
| Ting | 0.25 | 0.55 | 0.20 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{all 3 walk}) = 0.65 \times 0.40 \times 0.25 = 65/1000 = 13/200 = 0.065\) | M1, A1 | Ratios (eg 65:1000) are only penalised by 1 mark at first correct answer. Can be implied by correct answer. CAO; do not confuse with 0.65 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Rita by bus}) = 0.25\times(1– 0.15)\times(1– 0.20) = 17/100 = 0.17\) | M1, A1 | Can be implied by correct answer. CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\geq 2 \text{ cycle}) = 1 – [P(0 \text{ cycle}) + P(1 \text{ cycles})] = 1 – 0.702 = 0.298\) | B1, B1, M1, A1, (B1), (B1), (B1), (M1), (A1) | CAO at least 1 of these 3 terms or equivalent but allow a '×3'. CAO. Sum of 4 or 7 terms each a product of 3 probabilities but not '×3'. CAO. 1 – [sum of 4 terms each a product of 3 probabilities but not '×3']. CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{WW or CC}) = 0.585 + 0.070 = 0.655\) | B1, M1, A1 | CAO either. Sum of 2 terms each a product of 2 probabilities. CAO; or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{different}) = 1 – (b)(i) = 0.345\) | B1F | F on (b)(i) providing \(0 < p < 1\) |
## Part (a)(i)
$P(\text{all 3 walk}) = 0.65 \times 0.40 \times 0.25 = 65/1000 = 13/200 = 0.065$ | M1, A1 | Ratios (eg 65:1000) are only penalised by 1 mark at first correct answer. Can be implied by correct answer. CAO; do not confuse with 0.65 | 2
## Part (a)(ii)
$P(\text{Rita by bus}) = 0.25\times(1– 0.15)\times(1– 0.20) = 17/100 = 0.17$ | M1, A1 | Can be implied by correct answer. CAO | 2
## Part (a)(iii)
$P(2 \text{ cycle}) = 0.10 \times 0.45 \times (0.25 + 0.20) = 0.02025 + 0.10 \times (0.40 + 0.15) \times 0.55 = 0.03025 + (0.65 + 0.25) \times 0.45 \times 0.55 = 0.22275$ (0.27325)
$P(3 \text{ cycle}) = 0.10 \times 0.45\times 0.55 = 0.02475$
$P(\geq 2 \text{ cycle}) = P(2 \text{ cycle}) + P(3 \text{ cycle}) = 0.298$
or
$P(0 \text{ cycle}) = 0.90 \times 0.55 \times 0.45 = 0.22275$
$P(1 \text{ cycles}) = 0.10 \times 0.55 \times 0.45 = 0.02475 + 0.90 \times 0.45 \times 0.45 = 0.18225$ (0.47925) $+ 0.90 \times 0.55 \times 0.55 = 0.27225$
$P(\geq 2 \text{ cycle}) = 1 – [P(0 \text{ cycle}) + P(1 \text{ cycles})] = 1 – 0.702 = 0.298$ | B1, B1, M1, A1, (B1), (B1), (B1), (M1), (A1) | CAO at least 1 of these 3 terms or equivalent but allow a '×3'. CAO. Sum of 4 or 7 terms each a product of 3 probabilities but not '×3'. CAO. 1 – [sum of 4 terms each a product of 3 probabilities but not '×3']. CAO | 4
## Part (b)(i)
$P(\text{WW}) = (0.65 \times 0.90) = 0.585$
$P(\text{CC}) = (0.10 \times 0.70) = 0.070$
$P(\text{WW or CC}) = 0.585 + 0.070 = 0.655$ | B1, M1, A1 | CAO either. Sum of 2 terms each a product of 2 probabilities. CAO; or equivalent | 3
## Part (b)(ii)
$P(\text{different}) = 1 – (b)(i) = 0.345$ | B1F | F on (b)(i) providing $0 < p < 1$ | 1
---4 Each school-day morning, three students, Rita, Said and Ting, travel independently from their homes to the same school by one of three methods: walk, cycle or bus. The table shows the probabilities of their independent daily choices.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Walk & Cycle & Bus \\
\hline
Rita & 0.65 & 0.10 & 0.25 \\
\hline
Said & 0.40 & 0.45 & 0.15 \\
\hline
Ting & 0.25 & 0.55 & 0.20 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability that, on any given school-day morning:
\begin{enumerate}[label=(\roman*)]
\item all 3 students walk to school;
\item only Rita travels by bus to school;
\item at least 2 of the 3 students cycle to school.
\end{enumerate}\item Ursula, a friend of Rita, never travels to school by bus. The probability that:
Ursula walks to school when Rita walks to school is 0.9 ; Ursula cycles to school when Rita cycles to school is 0.7 .
Calculate the probability that, on any given school-day morning, Rita and Ursula travel to school by:
\begin{enumerate}[label=(\roman*)]
\item the same method;
\item different methods.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2010 Q4 [12]}}