| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Independence with three or more events |
| Difficulty | Standard +0.8 This S4 question requires understanding of independence conditions and careful probability calculations. Part (i) is straightforward application of independence. Part (ii) requires using the inclusion-exclusion principle with three independent events. Part (iii) is more challenging, requiring students to find bounds on P(A'∩B'∩C') by optimizing over valid probability constraints when B and C are not independent—this demands systematic reasoning about probability bounds rather than routine calculation. |
| Spec | 2.03a Mutually exclusive and independent events2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5\) | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.16 + 0.2 + 0.1 + 0.14 + 0.1 + 0.1 + 0.06\) | M2 | M1 for at least 4 correct. \(0.6+0.5+0.4-0.3-0.24-0.2+0.1=0.86\). M2A1. M1 if incorrect coefficient of \(P(A \cap B \cap C)\) used in otherwise correct formula. |
| \(= 0.86\) | A1 | |
| \(1 - \text{"0.86"}\) | ||
| \(0.14\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| greatest: \(P(A' \cap B \cap C') = 0.04\), \(P(A' \cap B \cap C) = 0.16\), \(P(A' \cap B' \cap C) = 0\) | M1 | For any of these, e.g. \(P(B \cap C) = 0.26\). Greatest: \(1-(0.6+0.5+0.4-0.3-0.24-0.26+0.1) = 0.2\) |
| least: \(P(A' \cap B \cap C') = 0.2\), \(P(A' \cap B \cap C) = 0\), \(P(A' \cap B' \cap C) = 0.16\) | M1 | For any of these, e.g. \(P(B \cap C) = 0.1\). Least: \(1-(0.6+0.5+0.4-0.3-0.24-0.1+0.1) = 0.04\) |
| greatest \(1-(0.16+0.2+0.04+0.14+0.1+0.16) = 0.2\) | M1A1 | M1 for fully correct method for either |
| least \(1-(0.16+0.2+0.2+0.14+0.1+0.16) = 0.04\) | A1 | |
| [5] |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5$ | B1 | |
| | **[1]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.16 + 0.2 + 0.1 + 0.14 + 0.1 + 0.1 + 0.06$ | M2 | M1 for at least 4 correct. $0.6+0.5+0.4-0.3-0.24-0.2+0.1=0.86$. M2A1. M1 if incorrect coefficient of $P(A \cap B \cap C)$ used in otherwise correct formula. |
| $= 0.86$ | A1 | |
| $1 - \text{"0.86"}$ | | |
| $0.14$ | A1 | |
| | **[5]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| greatest: $P(A' \cap B \cap C') = 0.04$, $P(A' \cap B \cap C) = 0.16$, $P(A' \cap B' \cap C) = 0$ | M1 | For any of these, e.g. $P(B \cap C) = 0.26$. Greatest: $1-(0.6+0.5+0.4-0.3-0.24-0.26+0.1) = 0.2$ |
| least: $P(A' \cap B \cap C') = 0.2$, $P(A' \cap B \cap C) = 0$, $P(A' \cap B' \cap C) = 0.16$ | M1 | For any of these, e.g. $P(B \cap C) = 0.1$. Least: $1-(0.6+0.5+0.4-0.3-0.24-0.1+0.1) = 0.04$ |
| greatest $1-(0.16+0.2+0.04+0.14+0.1+0.16) = 0.2$ | M1A1 | M1 for fully correct method for either |
| least $1-(0.16+0.2+0.2+0.14+0.1+0.16) = 0.04$ | A1 | |
| | **[5]** | |
---3 For events $A , B$ and $C$ it is given that $\mathrm { P } ( A ) = 0.6 , \mathrm { P } ( B ) = 0.5 , \mathrm { P } ( C ) = 0.4$ and $\mathrm { P } ( A \cap B \cap C ) = 0.1$. It is also given that events $A$ and $B$ are independent and that events $A$ and $C$ are independent.\\
(i) Find $\mathrm { P } ( B \mid A )$.\\
(ii) Given also that events $B$ and $C$ are independent, find $\mathrm { P } \left( A ^ { \prime } \cap B ^ { \prime } \cap C ^ { \prime } \right)$.\\
(iii) Given instead that events $B$ and $C$ are not independent, find the greatest and least possible values of $\mathrm { P } \left( A ^ { \prime } \cap B ^ { \prime } \cap C ^ { \prime } \right)$.
\hfill \mbox{\textit{OCR S4 2017 Q3 [10]}}