| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Multi-item selection from population |
| Difficulty | Standard +0.3 This is a straightforward probability question requiring basic combinatorics and conditional probability. Part (a) involves setting up P(both red) = C(r,2)/C(36,2) = 1/3 and simplifying to the given equation. Part (b) solves a simple quadratic. Parts (c)(i) and (c)(ii) apply standard probability formulas with no novel insight required. The multi-step nature and algebraic manipulation place it slightly above average, but all techniques are routine for S1 students. |
| Spec | 2.03a Mutually exclusive and independent events2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{r}{36} \times \frac{r-1}{35} = \frac{1}{3}\) → \(r(r-1) = 12 \times 35 = 420\) | M1 A1 M1 A1 | |
| (b) \(r^2 - r - 420 = 0\) → \((r - 21)(r + 20) = 0\) → \(r = 21\) | M1 M1 A1 | |
| (c) (i) \(P(\text{2 red}) = 3 \times \frac{21}{36} \times \frac{20}{35} = \frac{15}{34} \times \frac{15}{34}\) | M1 A1 M1 A1 | \(P(\text{3 red}) = \frac{21}{36} \times \frac{20}{35} \times \frac{19}{34} = \frac{19}{102}\) |
| (ii) \(P(\text{first is red | 2 or 3 red}) = \left(2 \times \frac{5}{34} + \frac{12}{102}\right) \div \frac{32}{51} = \frac{49}{64}\) or 0.766 | M1 A1 M1 A1 |
(a) $\frac{r}{36} \times \frac{r-1}{35} = \frac{1}{3}$ → $r(r-1) = 12 \times 35 = 420$ | M1 A1 M1 A1 |
(b) $r^2 - r - 420 = 0$ → $(r - 21)(r + 20) = 0$ → $r = 21$ | M1 M1 A1 |
(c) (i) $P(\text{2 red}) = 3 \times \frac{21}{36} \times \frac{20}{35} = \frac{15}{34} \times \frac{15}{34}$ | M1 A1 M1 A1 | $P(\text{3 red}) = \frac{21}{36} \times \frac{20}{35} \times \frac{19}{34} = \frac{19}{102}$ | M1 A1 M1 A1 | $P(\text{2 or 3 red}) = \frac{15}{34} + \frac{19}{102} = \frac{32}{51}$ or 0.627 | M1 A1 |
(ii) $P(\text{first is red | 2 or 3 red}) = \left(2 \times \frac{5}{34} + \frac{12}{102}\right) \div \frac{32}{51} = \frac{49}{64}$ or 0.766 | M1 A1 M1 A1 | Total: 17 marksSixteen cards have been lost from a pack, which therefore contains only 36 cards. Two cards are drawn at random from the pack. The probability that both cards are red is $\frac{1}{3}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $r$, the number of red cards in the pack, satisfies the equation
$$r(r - 1) = 420.$$ [4 marks]
\item Hence or otherwise find the value of $r$. [3 marks]
\item Find the probability that, when three cards are drawn at random from the pack,
\begin{enumerate}[label=(\roman*)]
\item at least two are red, [6 marks]
\item the first one is red given that at least two are red. [4 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [17]}}