## Part (i)
[Probability tree diagram with three stages]

| | | 
|---|---|
|First: 0.1 Hit, 0.9 Miss|Second: 0.2 Hit/0.8 Miss, 0.05 Hit/0.95 Miss|Third: 0.2 Hit/0.8 Miss, 0.05 Hit/0.95 Miss, 0.2 Hit/0.8 Miss, 0.05 Hit/0.95 Miss|

**Marks:** G1 | G1 | G1 | G1 | [4]

**Guidance:**
- G1: For first set of branches. All probabilities correct
- G1: For second set of branches (indep). All probabilities correct
- G1: For third set of branches (indep). All probabilities correct
- G1: For labels. All correct labels for 'Hit' and 'Miss', 'H' and 'M' etc. Condone omission of First, Second, Third
- Do not allow misreads here as all FT

## Part (ii)

**Answer (Method A):**

$$P(\text{Hits with at least one}) = 1 - P(\text{misses with all})$$
$$= 1 - (0.9 \times 0.95 \times 0.95) = 1 - 0.81225 = 0.18775$$

**ALTERNATIVE METHOD** only if there is an attempt to add 7 probabilities
At least three correct triple products
Attempt to add 7 triple products

**FURTHER ALTERNATIVE METHOD:**
$$0.1 + 0.9 \times 0.05$$
Above probability + $0.9 \times 0.95 \times 0.05$

**Marks:** M1* | M1* dep | A1 | M1 | M1 | A1 | M1 | M1 | A1 | [3]

**Guidance:**
- M1*: For 0.9 × 0.95 × 0.95
- M1* dep: For 1 – ans
- A1: CAO. 0.188 or better. Condone 0.1877. Allow 751/4000
- M1: (not necessarily correct triple products)
- M1: (not necessarily correct triple products)
- A1: CAO
- M1: 
- M1:
- A1: CAO

## Part (ii)

**Answer (Method B):**

$$P(\text{Hits with exactly one}) = (0.1 \times 0.8 \times 0.95) + (0.9 \times 0.05 \times 0.8) + (0.9 \times 0.95 \times 0.05)$$

$$= 0.076 + 0.036 + 0.04275 = \frac{19}{250} + \frac{9}{250} + \frac{171}{4000}$$

$$= \frac{619}{4000} = 0.15475$$

**Marks:** M1 | M1 | M1 | A1 | [4]

**Guidance:**
- M1: For two correct products
- M1: For all three correct products
- M1: For sum of all three correct products
- A1: CAO. Allow 0.155 or better

## Part (iii)
**Answer:**
$$P(\text{Hits with exactly one given hits with at least one}) = \frac{P(\text{Hits with exactly one and hits with at least one})}{P(\text{Hits with at least one})}$$

$$= \frac{0.15475}{0.18775} = 0.8242$$

**Marks:** M1 | M1 | A1 | [3]

**Guidance:**
- M1: For numerator FT
- M1: For denominator FT
- A1: CAO. Allow 0.824 or better or 619/751
- If answer to (B) > than answer to (A) then max M1M0A0

## Part (iv)
**Answer:**
$$P(\text{Hits three times overall}) = (0.1 \times 0.2 \times 0.2) + (0.9 \times 0.95 \times 0.05 \times 0.2 \times 0.2)$$

$$= 0.004 + 0.0016245$$

$$= 0.0056245$$

**Marks:** M1 | M1 | M1* Dep on both prev M1's | A1 | [4]

**Guidance:**
- M1: For 0.1 × 0.2 × 0.2 or 0.004 or 1/250
- M1: For 0.9 × 0.95 × 0.05 × 0.2 × 0.2
- M1* Dep on both prev M1's: For sum of both. With no extras
- A1: CAO. Allow 0.00562 or 0.00563 or 0.0056
- FT their tree for all three M marks, provided three terms in first product and six in second product. Last three probs must be 0.05 × 0.2 × 0.2 unless they extend their tree

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# Question 6 (repeat):

## Part (iii)
**Answer:**
Lower limit: $2.83 - (1.5 \times 0.88) = 1.51$

Upper limit: $3.71 + (1.5 \times 0.88) = 5.03$

Exactly one baby weighs less than 1.51 kg and none weigh over 5.03 kg so there is exactly one outlier.

**Marks:** B1 | B1 | E1* | [3] (Note: Different total from earlier listing)

**Guidance:** 
- B1: For 1.51 FT
- B1: For 5.03 FT
- E1*: Dep on their 1.51 and 5.03. For use of median ± 1.5 × IQR scores B0 B0 E0. No marks for ± 2 or 3 × IQR. In this part FT their values from (i)or (ii) if sensibly obtained but not from location ie 6.5, 19.5. Do not penalise over-specification as not the final answer. Do not allow unless FT leads to upper limit above 4.34 and lower limit between 1.39 and 2.50