| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with given score/outcome |
| Difficulty | Standard +0.8 This is a multi-part conditional probability question requiring systematic case analysis (counting matching pairs across three bags), application of conditional probability formula, and formal independence testing. While the individual probability calculations are straightforward, the combinatorial organization and the independence verification require careful reasoning beyond routine S1 exercises. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(2, N2, 2) = 1/4 \times 1 \times 1/7 = 1/28\) | M1 | Considering at least two options of 2s and 8s |
| \(P(8, 8, N8) = 1/4 \times 2/5 \times 3/7 = 3/70\) | M1 | Considering three options for the 8s |
| \(P(8, N8, 8) = 1/4 \times 3/5 \times 4/7 = 3/35\) | M1 | Summing their options if more than 3 in total |
| \(P(N8, 8, 8) = 3/4 \times 2/5 \times 4/7 = 6/35\) | B1 | One option correct |
| \(\Sigma = 47/140\ (0.336)\) | A1 [5] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(2, 2 \text{ given same}) = \dfrac{1/28}{47/140}\) | M1 | \(1/28\) in numerator of a fraction |
| \(= 5/47\ (0.106)\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X) = 47/140\) | M1 | Attempt to compare \(P(A \text{ and } B)\) with \(P(A) \times P(B)\) or using conditional probabilities |
| \(P(Y) = 1/4\) | ||
| \(P(X \text{ and } Y) = 1/28 \neq 47/140 \times 1/4\) | A1 | Legitimate correct answer |
| Not independent | [2] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(2, N2, 2) = 1/4 \times 1 \times 1/7 = 1/28$ | M1 | Considering at least two options of 2s and 8s |
| $P(8, 8, N8) = 1/4 \times 2/5 \times 3/7 = 3/70$ | M1 | Considering three options for the 8s |
| $P(8, N8, 8) = 1/4 \times 3/5 \times 4/7 = 3/35$ | M1 | Summing their options if more than 3 in total |
| $P(N8, 8, 8) = 3/4 \times 2/5 \times 4/7 = 6/35$ | B1 | One option correct |
| $\Sigma = 47/140\ (0.336)$ | A1 [5] | Correct answer |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(2, 2 \text{ given same}) = \dfrac{1/28}{47/140}$ | M1 | $1/28$ in numerator of a fraction |
| $= 5/47\ (0.106)$ | A1 [2] | Correct answer |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X) = 47/140$ | M1 | Attempt to compare $P(A \text{ and } B)$ with $P(A) \times P(B)$ or using conditional probabilities |
| $P(Y) = 1/4$ | | |
| $P(X \text{ and } Y) = 1/28 \neq 47/140 \times 1/4$ | A1 | Legitimate correct answer |
| Not independent | [2] | |7 Bag $A$ contains 4 balls numbered 2, 4, 5, 8. Bag $B$ contains 5 balls numbered 1, 3, 6, 8, 8. Bag $C$ contains 7 balls numbered $2,7,8,8,8,8,9$. One ball is selected at random from each bag.\\
(i) Find the probability that exactly two of the selected balls have the same number.\\
(ii) Given that exactly two of the selected balls have the same number, find the probability that they are both numbered 2 .\\
(iii) Event $X$ is 'exactly two of the selected balls have the same number'. Event $Y$ is 'the ball selected from bag $A$ has number 2'. Showing your working, determine whether events $X$ and $Y$ are independent or not.
\hfill \mbox{\textit{CAIE S1 2011 Q7 [9]}}