| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Probability distribution from conditional setup |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question requiring systematic enumeration of outcomes and basic probability calculations. Part (i) guides students through one case, part (ii) extends this to a simple distribution (X can only be 2, 3, or 4), and part (iii) applies standard conditional probability formula P(A|B) = P(A∩B)/P(B) with small numbers allowing direct counting. While requiring careful organization, it involves no novel insight and uses only fundamental S1 techniques, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) \(P(X = 3) = P(GRR) + P(RGR)\) | M1 | Mult 3 probs |
| \(\frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} + \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2}\) | M1 | Summing 2 options |
| \(= \frac{1}{3}\) AG | A1 3 | Correct working with appropriate justification and fraction sequencing |
| (ii) | X | 2 |
| Prob | \(\frac{1}{6}\) | \(\frac{1}{3}\) |
| \(P(X = 2) = P(RR) = \frac{2}{4} \times \frac{1}{3} = \frac{1}{6}\) | B1 | One correct prob other than (i) |
| \(P(X = 4) = 1 - \left(\frac{1}{6} + \frac{1}{3}\right) = \frac{1}{2}\) | ||
| Or P(GGR) + P(RGG) + P(GRG) | ||
| \(= \left(\frac{2}{4} \times \frac{1}{3} \times \frac{2}{1}\right) \times 3 = \frac{1}{2}\) | B1√ 3 | Second correct prob ft 1 – their previous 2 probs |
| (iii) \(P(\text{3 orange } | \text{ at least 2 O}) = \frac{P(3O)}{P(\text{at least 2O})}\) | M1 |
| \(P(\text{3 orange}) = P(OOO)\) | A1 | Correct unsimplified num of a fraction |
| \(= \frac{5}{6} \times \frac{4}{5} \times \frac{3}{5} = \frac{2}{7}\) | ||
| \(P(\text{at least 2O}) = P(YOO) + P(OYO) +\) | M1 | Attempt at P(at least 2O) sum 3 or 4 three-factor options |
| \(P(OOY) + \frac{2}{7}\) | ||
| \(= \frac{2}{6} \times \frac{5}{6} \times \frac{4}{5} + \frac{5}{6} \times \frac{2}{5} \times \frac{4}{5} + \frac{5}{6} \times \frac{4}{5} \times \frac{2}{5} + \frac{2}{7}\) | A1 | Correct unsimplified answer seen anywhere |
| \(= \frac{6}{7}\) | ||
| \(P(\text{3O} | \text{ at least 2O}) = \frac{2}{7} \div \frac{6}{7} = \frac{1}{3}\) (0.333) | A1 5 |
| Alternative 1 | ||
| 3 Orange = \(^5C_3\) | M1 | Attempt at combinations for 3 orange oe, not added |
| A1 | Correct unsimplified num of a fraction | |
| At least 2 Orange = \(^5C_2 \times ^2C_1 + ^5C_3\) | M1 | Attempt at combinations for at least 2 orange. Condone omission of \(+^2C_1\) |
| A1 | Correct unsimplified answer seen anywhere | |
| \(P(\text{3O} | \text{ at least 2O}) = \frac{^5C_3}{^5C_2 \times^2C_1 + ^5C_3} = \frac{1}{3}\) | A1 5 |
| Alternative 2 | ||
| No Yellow = \(^5C_0\) | M1 | Attempt at combinations for 0 yellow oe, not added |
| A1 | Correct unsimplified num of a fraction | |
| No more than 1 Yellow = \(^2C_1 + ^5C_0\) | M1 | Attempt at combinations for no more than 1 yellow. Condone omission of \(+2C0\) |
| A1 | Correct unsimplified answer seen anywhere | |
| \(P(\text{3O} | \text{ at least 2O}) = \frac{^5C_0}{^2C_1 + ^5C_0} = \frac{1}{3}\) | A1 5 |
| Misread – with replacement | ||
| MR–1 applied to first Accuracy Mark earned | M1 | Attempt at P(OOO) one three factor option oe not added |
| \(P(\text{3O}) = \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} = \frac{125}{343}\) | A1 | Correct unsimplified num of a fraction |
| \(P(\text{at least 2O}) = \frac{5}{7} \times \frac{5}{7} \times \frac{2}{7} \times ^3C_2 + \left(\frac{5}{7}\right)^3\) | M1 | Attempt at P(at least 2O) sum of 3 or 4 three factor options |
| A1 | Correct unsimplified seen anywhere | |
| \(P(\text{3O} | \text{ at least 2O}) = \frac{5}{11}\) | A1 4 max |
**(a) (i)** $P(X = 3) = P(GRR) + P(RGR)$ | M1 | Mult 3 probs
$\frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} + \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2}$ | M1 | Summing 2 options
$= \frac{1}{3}$ AG | A1 3 | Correct working with appropriate justification and fraction sequencing
**(ii)** | X | 2 | 3 | 4 | B1 | Values 2, 3, 4 only in table
| | Prob | $\frac{1}{6}$ | $\frac{1}{3}$ | $\frac{1}{2}$ | | Condone X=0,1 if P(X)=0 stated
| | | | |
$P(X = 2) = P(RR) = \frac{2}{4} \times \frac{1}{3} = \frac{1}{6}$ | B1 | One correct prob other than (i)
$P(X = 4) = 1 - \left(\frac{1}{6} + \frac{1}{3}\right) = \frac{1}{2}$ | |
Or P(GGR) + P(RGG) + P(GRG) | |
$= \left(\frac{2}{4} \times \frac{1}{3} \times \frac{2}{1}\right) \times 3 = \frac{1}{2}$ | B1√ 3 | Second correct prob ft 1 – their previous 2 probs
**(iii)** $P(\text{3 orange } | \text{ at least 2 O}) = \frac{P(3O)}{P(\text{at least 2O})}$ | M1 | Atttmept at P(OOO) one three-factor option, not added
$P(\text{3 orange}) = P(OOO)$ | A1 | Correct unsimplified num of a fraction
$= \frac{5}{6} \times \frac{4}{5} \times \frac{3}{5} = \frac{2}{7}$ |
$P(\text{at least 2O}) = P(YOO) + P(OYO) +$ | M1 | Attempt at P(at least 2O) sum 3 or 4 three-factor options
$P(OOY) + \frac{2}{7}$ |
$= \frac{2}{6} \times \frac{5}{6} \times \frac{4}{5} + \frac{5}{6} \times \frac{2}{5} \times \frac{4}{5} + \frac{5}{6} \times \frac{4}{5} \times \frac{2}{5} + \frac{2}{7}$ | A1 | Correct unsimplified answer seen anywhere
$= \frac{6}{7}$ |
$P(\text{3O} | \text{ at least 2O}) = \frac{2}{7} \div \frac{6}{7} = \frac{1}{3}$ (0.333) | A1 5 | Correct answer evaluated
**Alternative 1** |
3 Orange = $^5C_3$ | M1 | Attempt at combinations for 3 orange oe, not added
| A1 | Correct unsimplified num of a fraction
At least 2 Orange = $^5C_2 \times ^2C_1 + ^5C_3$ | M1 | Attempt at combinations for at least 2 orange. Condone omission of $+^2C_1$
| A1 | Correct unsimplified answer seen anywhere
$P(\text{3O} | \text{ at least 2O}) = \frac{^5C_3}{^5C_2 \times^2C_1 + ^5C_3} = \frac{1}{3}$ | A1 5 | Correct answer evaluated
**Alternative 2** |
No Yellow = $^5C_0$ | M1 | Attempt at combinations for 0 yellow oe, not added
| A1 | Correct unsimplified num of a fraction
No more than 1 Yellow = $^2C_1 + ^5C_0$ | M1 | Attempt at combinations for no more than 1 yellow. Condone omission of $+2C0$
| A1 | Correct unsimplified answer seen anywhere
$P(\text{3O} | \text{ at least 2O}) = \frac{^5C_0}{^2C_1 + ^5C_0} = \frac{1}{3}$ | A1 5 | Correct answer evaluated
**Misread – with replacement** |
MR–1 applied to first Accuracy Mark earned | M1 | Attempt at P(OOO) one three factor option oe not added
$P(\text{3O}) = \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} = \frac{125}{343}$ | A1 | Correct unsimplified num of a fraction
$P(\text{at least 2O}) = \frac{5}{7} \times \frac{5}{7} \times \frac{2}{7} \times ^3C_2 + \left(\frac{5}{7}\right)^3$ | M1 | Attempt at P(at least 2O) sum of 3 or 4 three factor options
| A1 | Correct unsimplified seen anywhere
$P(\text{3O} | \text{ at least 2O}) = \frac{5}{11}$ | A1 4 max | Answer evaluated7 A box contains 2 green apples and 2 red apples. Apples are taken from the box, one at a time, without replacement. When both red apples have been taken, the process stops. The random variable $X$ is the number of apples which have been taken when the process stops.\\
(i) Show that $\mathrm { P } ( X = 3 ) = \frac { 1 } { 3 }$.\\
(ii) Draw up the probability distribution table for $X$.
Another box contains 2 yellow peppers and 5 orange peppers. Three peppers are taken at random from the box without replacement.\\
(iii) Given that at least 2 of the peppers taken from the box are orange, find the probability that all 3 peppers are orange.
\hfill \mbox{\textit{CAIE S1 2014 Q7 [11]}}