| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Basic two-way table probability |
| Difficulty | Easy -1.3 This is a straightforward S1 conditional probability question requiring only reading values from a two-way table and applying basic probability formulas P(A|B) = P(A∩B)/P(B). Part (d) tests understanding of independence definition but involves simple arithmetic comparison. No problem-solving or conceptual insight needed—pure mechanical application of standard techniques. |
| Spec | 2.03a Mutually exclusive and independent events2.03d Calculate conditional probability: from first principles |
| Length of time employed, \(x\) years | Female | Male |
| \(x < 4\) | 9 | 16 |
| \(4 \leqslant x < 10\) | 14 | 20 |
| \(10 \leqslant x\) | 7 | 24 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{Female}) = \frac{30}{90}\) | B1 | \(\frac{30}{90}\) or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{Male} \mid < 4\text{ years}) = \frac{P(\text{Male} \cap < 4\text{ years})}{P(<4\text{ years})} = \frac{\frac{16}{90}}{\frac{16+9}{90}} = \frac{16}{25}\) | M1A1 | M1 for correct ratio expression with at least one correct probability substituted, or correct ratio of probabilities (num>denom is M0); A1 \(\frac{16}{25}\) or 0.64; correct answer scores 2 out of 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{Male} \mid < 10\text{ years}) = \frac{P(\text{Male} \cap < 10\text{ years})}{P(<10\text{ years})} = \frac{\frac{20+16}{90}}{\frac{9+16+14+20}{90}} = \frac{36}{59}\) | M1A1 | M1 for correct ratio expression with at least one correct probability; A1 \(\frac{36}{59}\) or condone awrt 0.610 (must be 3sf); correct answer scores 2 out of 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{Male} \mid <4\text{ years}) = \frac{16}{25}\), \(P(\text{Male}) = \frac{60}{90}\) or \(P(<4\text{ years}\mid\text{Male}) = \frac{16}{60}\), \(P(<4\text{ years}) = \frac{25}{90}\) or \(P(\text{Male} \cap <4\text{ years}) = \frac{16}{90}\), \(P(\text{Male}) = \frac{60}{90}\), \(P(<4\text{ years}) = \frac{25}{90}\) | M1 | 1st M1 for stating all required numerical probabilities for a correct test, which must be labelled; probabilities must be correct or correct ft from (b); if attempting first test, \(P(\text{Male}\mid<4\text{ years}) = \frac{16}{25}\) was found in (b) and need not be fully restated |
| \(P(M\mid<4) \neq P(M)\) or \(P(<4\mid M) \neq P(<4)\) or \(P(\text{Male}\cap<4\text{ years}) \neq P(M)\times P(<4)\) | M1 | 2nd M1 for use of a correct test; must see the product if attempting the 3rd test |
| So not independent. | A1 | A1 for correct test with all probabilities correct and correct conclusion; NB use of A and B throughout scores M0M0A0 unless A and B are explicitly defined |
## Question 3:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{Female}) = \frac{30}{90}$ | B1 | $\frac{30}{90}$ or exact equivalent |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{Male} \mid < 4\text{ years}) = \frac{P(\text{Male} \cap < 4\text{ years})}{P(<4\text{ years})} = \frac{\frac{16}{90}}{\frac{16+9}{90}} = \frac{16}{25}$ | M1A1 | M1 for correct ratio expression with at least one correct probability substituted, or correct ratio of probabilities (num>denom is M0); A1 $\frac{16}{25}$ or 0.64; correct answer scores 2 out of 2 |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{Male} \mid < 10\text{ years}) = \frac{P(\text{Male} \cap < 10\text{ years})}{P(<10\text{ years})} = \frac{\frac{20+16}{90}}{\frac{9+16+14+20}{90}} = \frac{36}{59}$ | M1A1 | M1 for correct ratio expression with at least one correct probability; A1 $\frac{36}{59}$ or condone awrt 0.610 (must be 3sf); correct answer scores 2 out of 2 |
### Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{Male} \mid <4\text{ years}) = \frac{16}{25}$, $P(\text{Male}) = \frac{60}{90}$ **or** $P(<4\text{ years}\mid\text{Male}) = \frac{16}{60}$, $P(<4\text{ years}) = \frac{25}{90}$ **or** $P(\text{Male} \cap <4\text{ years}) = \frac{16}{90}$, $P(\text{Male}) = \frac{60}{90}$, $P(<4\text{ years}) = \frac{25}{90}$ | M1 | 1st M1 for stating all required numerical probabilities for a correct test, which must be labelled; probabilities must be correct or correct ft from (b); if attempting first test, $P(\text{Male}\mid<4\text{ years}) = \frac{16}{25}$ was found in (b) and need not be fully restated |
| $P(M\mid<4) \neq P(M)$ **or** $P(<4\mid M) \neq P(<4)$ **or** $P(\text{Male}\cap<4\text{ years}) \neq P(M)\times P(<4)$ | M1 | 2nd M1 for use of a correct test; must see the product if attempting the 3rd test |
| So not independent. | A1 | A1 for correct test with all probabilities correct **and** correct conclusion; NB use of A and B throughout scores M0M0A0 unless A and B are explicitly defined |
---\begin{enumerate}
\item A company employs 90 administrators. The length of time that they have been employed by the company and their gender are summarised in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
Length of time employed, $x$ years & Female & Male \\
\hline
$x < 4$ & 9 & 16 \\
\hline
$4 \leqslant x < 10$ & 14 & 20 \\
\hline
$10 \leqslant x$ & 7 & 24 \\
\hline
\end{tabular}
\end{center}
One of the 90 administrators is selected at random.\\
(a) Find the probability that the administrator is female.\\
(b) Given that the administrator has been employed by the company for less than 4 years, find the probability that this administrator is male.\\
(c) Given that the administrator has been employed by the company for less than 10 years, find the probability that this administrator is male.\\
(d) State, with a reason, whether or not the event 'selecting a male' is independent of the event 'selecting an administrator who has been employed by the company for less than 4 years'.\\
\hfill \mbox{\textit{Edexcel S1 2015 Q3 [8]}}