# Question 2:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Inter-quartile range $= 18.1 - 17.8 = 0.3$ | B1 | |
| Lower limit $17.8 - (1.5 \times 0.3)$ $(= 17.35)$; No outliers at lower end. | M1, A1 | dep on 17.35 — FT their IQR for M marks only. Allow 'No values below 17.35 for first A1'. Allow 'Lower limit $= 17.35$ so no outliers (at lower end)'. Watch for use of median giving 17.45 which gets M0A0. You must be convinced that comments about no outliers refer to lower tail only. |
| Upper limit $18.1 + (1.5 \times 0.3)$ $(= 18.55)$; (Max is 18.6) so at least one outlier at upper end. | M1, A1 | dep on 18.55 — Allow 'At least one value above 18.55' for second A1. Allow 'any value above 18.55 is an outlier' so at least one outlier. Do not allow 'There MAY be one outlier'. Condone 'one outlier'. Condone 'there are outliers'. Watch for use of median giving 18.35 which gets M0A0. |

**Total: [5]**

## Part (ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(A) = P(\text{All 3 have orange centres}) = \frac{7}{20} \times \frac{6}{19} \times \frac{5}{18} = \frac{7}{228}$ | M1 | For $\frac{7}{20}\times$ |
| | M1 | For product of correct three fractions without extra terms |
| $= 0.0307\ (0.030702)$ | A1 | CAO — Allow full marks for fully simplified fractional answers |

**Alternative scheme** $^7C_3/^{20}C_3 = 35/1140 = 7/228 = 0.0307$: M1 for either term in correct position in a fraction, M1 for correct fraction, A1 CAO

| $P(B) = P(\text{All 3 have same centres})$ $= \left(\frac{7}{20}\times\frac{6}{19}\times\frac{5}{18}\right)+\left(\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\right)+\left(\frac{4}{20}\times\frac{3}{19}\times\frac{2}{18}\right)+\left(\frac{3}{20}\times\frac{2}{19}\times\frac{1}{18}\right)$ | M1 | For at least two correct triple products or fractions or decimals |
|--------|-------|----------|
| $= 0.0307 + 0.0175 + 0.0035 + 0.0009$ | M1 | For sum of all four correct |
| $= 0.0526 = \frac{1}{19}\ (0.052632)$ $\left(=\frac{7}{228}+\frac{1}{57}+\frac{1}{285}+\frac{1}{1140}\right)$ | A1 | CAO — Allow 0.053 or anything which rounds to 0.053 with working |

**Total: [6]**

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(A\mid B) = \dfrac{0.0307..}{0.0526..}$ | M1 | For their $A$ divided by their $B$ — Allow 0.584 from $\frac{0.0307}{0.0526}$ |
| $= 0.583\ (= 0.58333)$ | A1 | FT their answers to (ii) provided answer $< 1$ — Allow $\frac{7}{12}$ |
| $P(B\mid A) = 1$ | B1 | CAO |

**Total: [3]**

## Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{All have orange centres}) = 0.0307^2 = 0.00094$ or $\frac{49}{51984}$ $(= 0.00094260)$ | M1 | For their $0.0307^2$ — Allow $9.4 \times 10^{-4}$, condone 0.0009 or $9 \times 10^{-4}$ |
| | A1 | FT |

**Total: [2]**

## Part (v)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Has to select} > 2) = 1 - P(\text{Has to select} \leq 2)$ $= 1 - \left(\frac{14}{20}+\left(\frac{6}{20}\times\frac{14}{19}\right)\right) = 1-(0.7+0.221) = 1-0.921$ | M1 | For $\left(\frac{6}{20}\times\frac{14}{19}\right)$ — For any of the methods below allow SC2 for $1-0.079 = 0.921$ or $1 - 3/38 = 35/38$ o.e. as final answer. This is $1 - P(C' + CC')$ |
| | M1 | For $1 -$ sum of both |
| $= 0.079\ (=0.078947)$ | A1 | CAO |

**Total: [3]**

**OR:** $P(\text{Has to select} > 2) = P(\text{First 2 both cherry}) = \left(\frac{6}{20}\times\frac{5}{19}\right)$

| | M2 | For whole product — Without extra terms added. M1 if multiplied by $k/18$ only where $0<k<18$. This is $P(CC)$. |
|--------|-------|----------|
| $= 0.079 = \frac{3}{38}$ | A1 | CAO |

**OR:** $1-(P(\text{0 cherries})+P(\text{1 cherry}))$
$= 1-\left(\frac{14}{20}\times\frac{13}{19}+\left(\frac{6}{20}\times\frac{14}{19}\right)+\left(\frac{14}{20}\times\frac{6}{19}\right)\right)$
$= 1-(0.4789+0.2211+0.2211) = 1-0.9209$

| | M1 | For any term — This is $1-P(C'C'+CC'+C'C)$ |
|--------|-------|----------|
| | M1 | For $1 -$ sum of all three |
| $= 0.079$ | A1 | CAO |

**OR** (full enumeration method):
$\left(\frac{6}{20}\times\frac{5}{19}\times\frac{14}{18}\right)+\left(\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{14}{17}\right)+\ldots$

| | M1 | For any term — This is $P(CCC'+CCCC'+CCCCC'+CCCCCC')$ |
|--------|-------|----------|
| $= \frac{7}{114}+\frac{14}{969}+\frac{7}{2584}+\frac{7}{19380}+\frac{1}{38760}$ | M1 | For sum of all five terms (all correct) |
| $= 0.079$ | A1 | CAO |