| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Incomplete two-way table completion |
| Difficulty | Moderate -0.3 This is a straightforward conditional probability question using a two-way table setup. Parts (a)-(c) require basic application of P(A|B) = P(A∩B)/P(B) with given proportions, and part (d) involves a simple combination calculation. The arithmetic is clean (fractions of 5 and 10) and no novel insight is required—just systematic application of standard S1 techniques. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(M \mid L) = \frac{P(M \cap L)}{P(L)} = \frac{\frac{3}{5} \times \frac{1}{5}}{\frac{3}{10}}\) | M1 | Fully correct ratio or correct ratio expression with at least one correct probability; if numerator \(>\) denominator then M0 |
| \(= \mathbf{0.40}\) (o.e.) | A1 | 0.40 or any exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = P(L\mid F) = \frac{P(L \cap F)}{P(F)} = \frac{\frac{3}{10} - \frac{3}{5}\times\frac{1}{5}}{1-\frac{3}{5}}\) or \(\frac{3}{5}\times\frac{1}{5} + \left(1-\frac{3}{5}\right)\times x = \frac{3}{10}\) | M1 | Equation for \(x\) with at least 2 of \(\left(\frac{3}{5}\times\frac{1}{5}\right)\) or \(\frac{3}{10}\) or \(\left(1-\frac{3}{5}\right)\) correct; \(\frac{\frac{2}{5}\times\frac{3}{10}}{\frac{2}{5}}\) is M0 |
| \(x = \frac{0.3-0.12}{0.40}\) or \(0.4x = 0.3 - 0.12\) | M1 | Fully correct expression for \(x\) |
| \(x = \mathbf{0.45}\) (o.e.) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(M \cap R) = 0.6 - P(M \cap L)\) or \(0.6 \times (1-0.2)\) | M1 | Correct expression with 0.6; follow through their \(P(M \cap L) = 0.12\) |
| \(= \mathbf{0.48}\) (o.e.) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{one left handed and other right handed}) = 2 \times \frac{3}{10} \times \frac{7}{10} = \frac{21}{50}\) or \(\mathbf{0.42}\) | M1, A1 | M1 for fully correct expression including the 2; allow \(1-0.3\) instead of 0.7 |
# Question 7:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(M \mid L) = \frac{P(M \cap L)}{P(L)} = \frac{\frac{3}{5} \times \frac{1}{5}}{\frac{3}{10}}$ | M1 | Fully correct ratio or correct ratio expression with at least one correct probability; if numerator $>$ denominator then M0 |
| $= \mathbf{0.40}$ (o.e.) | A1 | 0.40 or any exact equivalent |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = P(L\mid F) = \frac{P(L \cap F)}{P(F)} = \frac{\frac{3}{10} - \frac{3}{5}\times\frac{1}{5}}{1-\frac{3}{5}}$ or $\frac{3}{5}\times\frac{1}{5} + \left(1-\frac{3}{5}\right)\times x = \frac{3}{10}$ | M1 | Equation for $x$ with at least 2 of $\left(\frac{3}{5}\times\frac{1}{5}\right)$ or $\frac{3}{10}$ or $\left(1-\frac{3}{5}\right)$ correct; $\frac{\frac{2}{5}\times\frac{3}{10}}{\frac{2}{5}}$ is M0 |
| $x = \frac{0.3-0.12}{0.40}$ or $0.4x = 0.3 - 0.12$ | M1 | Fully correct expression for $x$ |
| $x = \mathbf{0.45}$ (o.e.) | A1 | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(M \cap R) = 0.6 - P(M \cap L)$ or $0.6 \times (1-0.2)$ | M1 | Correct expression with 0.6; follow through their $P(M \cap L) = 0.12$ |
| $= \mathbf{0.48}$ (o.e.) | A1 | |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{one left handed and other right handed}) = 2 \times \frac{3}{10} \times \frac{7}{10} = \frac{21}{50}$ or $\mathbf{0.42}$ | M1, A1 | M1 for fully correct expression including the 2; allow $1-0.3$ instead of 0.7 |7. In a large college, $\frac { 3 } { 5 }$ of the students are male, $\frac { 3 } { 10 }$ of the students are left handed and $\frac { 1 } { 5 }$ of the male students are left handed.
A student is chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Given that the student is left handed, find the probability that the student is male.
\item Given that the student is female, find the probability that she is left handed.
\item Find the probability that the randomly chosen student is male and right handed.
Two students are chosen at random.
\item Find the probability that one student is left handed and one is right handed.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q7 [9]}}