| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Two-Way Table to Probability |
| Difficulty | Moderate -0.8 This is a straightforward two-way table problem requiring basic probability calculations and an independence check. Part (i) uses simple addition of probabilities (or counting), and part (ii) applies the standard definition P(A∩B) = P(A)×P(B). Both parts involve routine manipulation of given data with no conceptual challenges beyond AS-level probability fundamentals. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(F) = \frac{12}{30}\ (0.4)\) | B1 | \(\frac{12}{30}\) or \(\frac{16}{30}\) or \(\frac{5}{30}\) seen |
| or \(P(W) = \frac{16}{30}\ (0.533)\) | M1 | Valid attempt to find \(P(F\) or \(W)\) |
| or \(P(M \cap W') = \frac{5}{30}\ (0.167)\) | ||
| \((F\ \text{or}\ W) = \frac{13}{30}+\frac{3}{30}+\frac{9}{30}\) | A1 | Correct unsimplified expression |
| or \(1-\frac{5}{30}\) or \(\frac{12}{30}+\frac{16}{30}-\frac{3}{30}\) | ||
| \(= \frac{5}{6}\ (0.833)\) | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(M) = 18/30\ (0.6)\), \(P(W) = 16/30\ (0.533)\), \(P(M) \times P(W) = 8/25\ (0.32)\) | M1 | Valid attempt to find \(P(M)\), \(P(W)\) and \(P(M) \times P(W)\) |
| \(P(M\ \text{and}\ W) = 13/30\ (0.433) \neq 8/25\ (0.32)\) | A1 | \(P(M\ \text{and}\ W) = 13/30 \neq 8/25\) and correct conclusion |
| not independent | ||
| OR | ||
| \(P(M | W) = \frac{P(M\ \text{and}\ W)}{P(W)} = \frac{13/30}{16/30} = \frac{13}{16}\ (0.813)\) | M1 |
| \(\neq \frac{18}{30} = P(M)\) | A1 | \(\frac{13}{16} \neq \frac{18}{30} = P(M)\) |
| not independent | ||
| OR | ||
| \(P(W | M) = \frac{P(M\ \text{and}\ W)}{P(M)} = \frac{13/30}{18/30} = \frac{13}{18}\) | M1 |
| \(\neq \frac{16}{30} = P(W)\) | A1 | \(\frac{13}{16} \neq \frac{18}{30} = P(M)\) |
| not independent | [2] |
## Question 2:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(F) = \frac{12}{30}\ (0.4)$ | B1 | $\frac{12}{30}$ or $\frac{16}{30}$ or $\frac{5}{30}$ seen |
| or $P(W) = \frac{16}{30}\ (0.533)$ | M1 | Valid attempt to find $P(F$ or $W)$ |
| or $P(M \cap W') = \frac{5}{30}\ (0.167)$ | | |
| $(F\ \text{or}\ W) = \frac{13}{30}+\frac{3}{30}+\frac{9}{30}$ | A1 | Correct unsimplified expression |
| or $1-\frac{5}{30}$ or $\frac{12}{30}+\frac{16}{30}-\frac{3}{30}$ | | |
| $= \frac{5}{6}\ (0.833)$ | A1 [4] | Correct answer |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(M) = 18/30\ (0.6)$, $P(W) = 16/30\ (0.533)$, $P(M) \times P(W) = 8/25\ (0.32)$ | M1 | Valid attempt to find $P(M)$, $P(W)$ and $P(M) \times P(W)$ |
| $P(M\ \text{and}\ W) = 13/30\ (0.433) \neq 8/25\ (0.32)$ | A1 | $P(M\ \text{and}\ W) = 13/30 \neq 8/25$ and correct conclusion |
| not independent | | |
| **OR** | | |
| $P(M|W) = \frac{P(M\ \text{and}\ W)}{P(W)} = \frac{13/30}{16/30} = \frac{13}{16}\ (0.813)$ | M1 | Valid attempt to find $P(M\ \text{and}\ W)$, $P(W)$ and $P(M\ \text{and}\ W) \div P(W)$ |
| $\neq \frac{18}{30} = P(M)$ | A1 | $\frac{13}{16} \neq \frac{18}{30} = P(M)$ |
| not independent | | |
| **OR** | | |
| $P(W|M) = \frac{P(M\ \text{and}\ W)}{P(M)} = \frac{13/30}{18/30} = \frac{13}{18}$ | M1 | Valid attempt to find $P(M\ \text{and}\ W)$, $P(M)$ and $P(M\ \text{and}\ W) \div P(M)$ |
| $\neq \frac{16}{30} = P(W)$ | A1 | $\frac{13}{16} \neq \frac{18}{30} = P(M)$ |
| not independent | [2] | |
---2 In a group of 30 teenagers, 13 of the 18 males watch 'Kops are Kids' on television and 3 of the 12 females watch 'Kops are Kids'.\\
(i) Find the probability that a person chosen at random from the group is either female or watches 'Kops are Kids' or both.\\
(ii) Showing your working, determine whether the events 'the person chosen is male' and 'the person chosen watches Kops are Kids' are independent or not.
\hfill \mbox{\textit{CAIE S1 2011 Q2 [6]}}