| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Probability distribution from conditional setup |
| Difficulty | Moderate -0.3 This is a straightforward conditional probability question using basic counting principles. Parts (i)-(iii) involve simple combinations from a finite set with clear structure (2 ones, 4 threes, 3 fives), and part (iv) requires organizing already-calculated probabilities into a table. While it requires careful counting and understanding of conditional probability, it's more routine than average A-level questions that typically involve calculus or algebraic manipulation alongside probability. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(\text{same}) = P(1,1) + P(3,3) + P(5,5) = \frac{2}{9} \cdot \frac{1}{8} + \frac{4}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{2}{8} = \frac{5}{18} (0.278)\) | M1 | Summing 3 two-factor options |
| M1 | Multiplying terms by one less in the numerator or denominator | |
| A1 3 | Correct answer | |
| Alt. method: \(\frac{2C2 + 4C2 + 3C2}{9C2} = \frac{2 \times 1 + 3 \times 4 + 2 \times 3}{9C2 \times 2}\) oe | M1 for numerator, M1 for denominator, A1 correct answer | |
| (ii) \(P(5,5) + P(5,5) = \frac{3}{9} \cdot \frac{6}{8} + \frac{6}{9} \cdot \frac{3}{8} = \frac{36}{72} = \frac{1}{2}\) or 0.5 | M1 | Mult 2 probs whose numerators sum to 9 o.e. |
| M1 | Summing 2 options or mult by 2 (may be 4 options) | |
| A1 3 | Correct answer | |
| Alt. method: \(\frac{6C1 \times 3C1(\times2)}{9C2(\times2)}\) oe | M1 for numerator, M1 for denominator, A1 correct answer | |
| (iii) \(P(5 \cap \bar{5}) = \frac{3}{9} \times \frac{6}{8} = \frac{1}{4}\) | M1 | Attempt at P(5 and not 5) seen as numerator or denominator of a fraction |
| \(P(\bar{5}) = \frac{1}{4} + \frac{6}{9} \times \frac{5}{8} = 48/72 = 0.6666\) | M1 | Attempt at P(not 5) sum of 2 two-factor terms seen anywhere |
| \(P(5 | \bar{5}_s) = \frac{1/4}{48/72} = \frac{3}{8} = 0.375\) | A1 |
| A1 4 | Correct answer | |
| (iv) | x | 0 |
| P(X = x) | 5/12 | 1/2 |
| \(P(0) = P(5,\bar{5}) = \frac{6}{9} \times \frac{5}{8} = 30/72 (5/12)\) | B1 | Values 0, 1, 2 seen in table with at least 1 prob |
| B1 | Correct P(0) unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(2) = 6/72 (1/12) (0.0833)\) from part (i) | B1 ft 3 | If \(x=0,1,2(3)\) ft \(\Sigma p = 1\), no –ve values, all probabilities \(< 1\) |
(i) $P(\text{same}) = P(1,1) + P(3,3) + P(5,5) = \frac{2}{9} \cdot \frac{1}{8} + \frac{4}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{2}{8} = \frac{5}{18} (0.278)$ | M1 | Summing 3 two-factor options
| M1 | Multiplying terms by one less in the numerator or denominator
| A1 3 | Correct answer
Alt. method: $\frac{2C2 + 4C2 + 3C2}{9C2} = \frac{2 \times 1 + 3 \times 4 + 2 \times 3}{9C2 \times 2}$ oe | | M1 for numerator, M1 for denominator, A1 correct answer
(ii) $P(5,5) + P(5,5) = \frac{3}{9} \cdot \frac{6}{8} + \frac{6}{9} \cdot \frac{3}{8} = \frac{36}{72} = \frac{1}{2}$ or 0.5 | M1 | Mult 2 probs whose numerators sum to 9 o.e.
| M1 | Summing 2 options or mult by 2 (may be 4 options)
| A1 3 | Correct answer
Alt. method: $\frac{6C1 \times 3C1(\times2)}{9C2(\times2)}$ oe | | M1 for numerator, M1 for denominator, A1 correct answer
(iii) $P(5 \cap \bar{5}) = \frac{3}{9} \times \frac{6}{8} = \frac{1}{4}$ | M1 | Attempt at P(5 and not 5) seen as numerator or denominator of a fraction
$P(\bar{5}) = \frac{1}{4} + \frac{6}{9} \times \frac{5}{8} = 48/72 = 0.6666$ | M1 | Attempt at P(not 5) sum of 2 two-factor terms seen anywhere
$P(5 | \bar{5}_s) = \frac{1/4}{48/72} = \frac{3}{8} = 0.375$ | A1 | Correct P($\bar{5}$) as numerator or denominator in fraction
| A1 4 | Correct answer
(iv) | x | 0 | 1 | 2 |
|---|---|---|---|
| P(X = x) | 5/12 | 1/2 | 1/12 |
$P(0) = P(5,\bar{5}) = \frac{6}{9} \times \frac{5}{8} = 30/72 (5/12)$ | B1 | Values 0, 1, 2 seen in table with at least 1 prob
| B1 | Correct P(0) unsimplified
$P(1) = 0.5$ from part (ii)
$P(2) = 6/72 (1/12) (0.0833)$ from part (i) | B1 ft 3 | If $x=0,1,2(3)$ ft $\Sigma p = 1$, no –ve values, all probabilities $< 1$7 Dayo chooses two digits at random, without replacement, from the 9-digit number 113333555.\\
(i) Find the probability that the two digits chosen are equal.\\
(ii) Find the probability that one digit is a 5 and one digit is not a 5 .\\
(iii) Find the probability that the first digit Dayo chose was a 5, given that the second digit he chose is not a 5 .\\
(iv) The random variable $X$ is the number of 5s that Dayo chooses. Draw up a table to show the probability distribution of $X$.
\hfill \mbox{\textit{CAIE S1 2013 Q7 [13]}}