| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Person selected from combined populations |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question using basic probability rules and Bayes' theorem. It requires only direct application of formulas with clearly stated percentages and populations, involving simple arithmetic with no conceptual challenges or problem-solving insight needed. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X) = \frac{20}{28}\left(\frac{5}{7}\right)(0.714), 71.4\%\) | B1 (1) | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(F) = \frac{20}{28}\times\frac{1}{4}\times\frac{8}{28}\times\frac{6}{10} = \frac{7}{20}\) | M1 | Summing two 2-factor probs created by one of \(\frac{1}{4}\) or \(\frac{3}{4}\) multiplied by \(\frac{20}{28}\) or \(\frac{8}{28}\); added to \(\frac{4}{10}\) or \(\frac{6}{10} \times\) altn population prob |
| A1 (2) | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X\mid F) = \frac{5/28}{7/20} = \frac{25}{49}\ (0.510)\) | M1 | Their unsimplified country X probability (\(\frac{5}{28}\)) as num or denom of a fraction, or (their fair hair population) \(\div\) (total fair hair pop) |
| A1 (2) | Correct answer |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X) = \frac{20}{28}\left(\frac{5}{7}\right)(0.714), 71.4\%$ | B1 (1) | oe |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(F) = \frac{20}{28}\times\frac{1}{4}\times\frac{8}{28}\times\frac{6}{10} = \frac{7}{20}$ | M1 | Summing two 2-factor probs created by one of $\frac{1}{4}$ or $\frac{3}{4}$ multiplied by $\frac{20}{28}$ or $\frac{8}{28}$; added to $\frac{4}{10}$ or $\frac{6}{10} \times$ altn population prob |
| | A1 (2) | Correct answer |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X\mid F) = \frac{5/28}{7/20} = \frac{25}{49}\ (0.510)$ | M1 | Their unsimplified country X probability ($\frac{5}{28}$) as num or denom of a fraction, or (their fair hair population) $\div$ (total fair hair pop) |
| | A1 (2) | Correct answer |
---2 In country $X , 25 \%$ of people have fair hair. In country $Y , 60 \%$ of people have fair hair. There are 20 million people in country $X$ and 8 million people in country $Y$. A person is chosen at random from these 28 million people.\\
(i) Find the probability that the person chosen is from country $X$.\\
(ii) Find the probability that the person chosen has fair hair.\\
(iii) Find the probability that the person chosen is from country $X$, given that the person has fair hair.
\hfill \mbox{\textit{CAIE S1 2015 Q2 [5]}}