| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with algebraic expressions |
| Difficulty | Standard +0.3 This is a standard S1 Venn diagram question requiring systematic application of probability rules (sum to 1, independence definition P(B∩C)=P(B)P(C)) and conditional probability formula. While multi-part with algebraic manipulation, it follows a predictable template with no novel insight required—slightly easier than average A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(p = P(B \cap C) = P(B) × P(C) = 0.6 × 0.25 = 0.15\) | M1 |
| \(q = [P(C) - p] = 0.10\) | A1 | For \(q = 0.10\) (both correct 2/2) |
| (b) | \(r = 1 - 0.08 - [P(B) + q] = 1 - 0.08 - 0.6 - 0.1\) (o.e.) or \(1 - 0.08 - (0.6 + 0.25 - p)\) | M1 |
| \(= 0.22\) | A1cao | For \(r = 0.22\) (correct ans only 2/2) |
| (c) | \(s = [P(A) - r] = 0.28\) | B1ft |
| \(t = [P(B) - p - s\) or use \(P(B \cap C') - s = 0.6 × 0.75 - "0.28"] = 0.17\) | B1ft | 2nd B1ft for \(t = 0.17\) or \(0.6 -\) their "\(0.15"\) \(-\) their "\(0.28"\) |
| (d) | \(P(A) × P(B) = 0.5 × 0.6 = 0.3\) which is not equal to \(s\) (= 0.28) | M1 |
| So \(A\) and \(B\) are not independent | A1 | Dep. on M1 being earned and clear statement that \(A\) and \(B\) are not independent |
| (e) | \(P(A \cup C)\) or \([P(A) + P(C)]\) or \([r + s + p + q]\) | M1, A1ft |
| \(= \frac{\text{("0.28"+"0.15") or (0.6 - "0.17")}}{0.5 + 0.25} = \frac{43}{75}\) | 1st A1ft for correct ratio or ft using their values in numerator but correct denominator. 2nd A1 for \(\frac{43}{75}\) or accept awrt 0.573 |
(a) | $p = P(B \cap C) = P(B) × P(C) = 0.6 × 0.25 = 0.15$ | M1 | For a correct expression (using independence) for $p$ or 0.15 |
| $q = [P(C) - p] = 0.10$ | A1 | For $q = 0.10$ (both correct 2/2) |
(b) | $r = 1 - 0.08 - [P(B) + q] = 1 - 0.08 - 0.6 - 0.1$ (o.e.) or $1 - 0.08 - (0.6 + 0.25 - p)$ | M1 | For a correct expression for $r$ using $P(B \cup C)$. Can it their $q \in [0, 0.32]$ |
| $= 0.22$ | A1cao | For $r = 0.22$ (correct ans only 2/2) |
(c) | $s = [P(A) - r] = 0.28$ | B1ft | 1st B1ft for $s = 0.28$ or $0.5 -$ their "$0.22$" |
| $t = [P(B) - p - s$ or use $P(B \cap C') - s = 0.6 × 0.75 - "0.28"] = 0.17$ | B1ft | 2nd B1ft for $t = 0.17$ or $0.6 -$ their "$0.15"$ $-$ their "$0.28"$ |
(d) | $P(A) × P(B) = 0.5 × 0.6 = 0.3$ which is not equal to $s$ (= 0.28) | M1 | For a correct $P(A) × P(B) = 0.5 × 0.6$ or 0.3 and a clear comparison with their $s(≠ 0.3)$ |
| So $A$ and $B$ are not independent | A1 | Dep. on M1 being earned and clear statement that $A$ and $B$ are not independent |
(e) | $P(A \cup C)$ or $[P(A) + P(C)]$ or $[r + s + p + q]$ | M1, A1ft | M1 for a correct ratio expression of probs: num. $<$ den. Allow $1 - (0.08+\text{their}$ "$r$") on den. Any sight of multiplication on the numerator e.g. $0.6 × 0.75$ is M0 |
| $= \frac{\text{("0.28"+"0.15") or (0.6 - "0.17")}}{0.5 + 0.25} = \frac{43}{75}$ | | 1st A1ft for correct ratio or ft using their values in numerator but correct denominator. 2nd A1 for $\frac{43}{75}$ or accept awrt 0.573 |\begin{enumerate}
\item The Venn diagram shows three events $A , B$ and $C$, where $p , q , r , s$ and $t$ are probabilities.\\
\includegraphics[max width=\textwidth, alt={}, center]{319667e7-3f8b-4a33-8fc5-ef72154d1421-10_647_972_306_488}\\
(b) Find the value of $r$.\\
(c) Hence write down the value of $s$ and the value of $t$.\\
(d) State, giving a reason, whether or not the events $A$ and $B$ are independent.\\
(e) Find $\mathrm { P } ( B \mid A \cup C )$.\\
$\mathrm { P } ( A ) = 0.5 , \mathrm { P } ( B ) = 0.6$ and $\mathrm { P } ( C ) = 0.25$ and the events $B$ and $C$ are independent.\\
(a) Find the value of $p$ and the value of $q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2017 Q3 [11]}}