## Question 7(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $[(0.6)^4 \times 0.4 =]\ 0.0518[4],\ \frac{162}{3125}$ | B1 | |

**Total: 1 mark**

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## Question 7(b):

**Method 1:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X \leq 7) - P(X \leq 2) =]\ (1 - 0.6^7) - (1 - 0.6^2)$ | M1 | $(1-p^7)-(1-p^2)$ or $p^2 - p^7$ seen, $0 < p < 1$ |
| $[= 0.36 - 0.02799]$ $= 0.332[0\ldots],\ \frac{25938}{78125}$ | A1 | If M0 awarded **SC B1** $0.3320064$ or $\frac{25938}{78125}$ CAO |

**Method 2:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X=3,4,5,6,7)=]\ 0.4\times0.6^2 + 0.4\times0.6^3 + 0.4\times0.6^4 + 0.4\times0.6^5 + 0.4\times0.6^6$ | M1 | $(1-p)\times p^2 + (1-p)\times p^3 + (1-p)\times p^4 + (1-p)\times p^5 + (1-p)\times p^6$ seen, $0 < p < 1$ |
| $= 0.332[0\ldots],\ \frac{25938}{78125}$ | A1 | If M0 awarded **SC B1** $0.3320064$ or $\frac{25938}{78125}$ CAO |

**Method 3 – geometric series:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X=3,4,5,6,7)=]\ \frac{0.144(1-0.6^5)}{1-0.6\ or\ 0.4}$ | M1 | $\frac{0.144(1-p^5)}{1-p}$ seen, $0 < p < 1$ |
| $= 0.332[0\ldots],\ \frac{25938}{78125}$ | A1 | If M0 awarded **SC B1** $0.3320064$ or $\frac{25938}{78125}$ CAO |

**Total: 2 marks**

## Question 7(c):

**Method 1**

| Answer | Mark | Guidance |
|--------|------|----------|
| 2 correct unsimplified outcomes identified | M1 | Condone not identified but not incorrectly identified |
| 2nd attempt: $(0.4)^2$ $[=0.16]$ | M1 | Add values for 4 identified correct scenarios. Condone adding values of 2nd, 3rd and 4th attempts only. No incorrect scenarios. |
| 3rd attempt: $(0.4)^2(0.6) \times (2 \text{ or } ^2C_1)$ $[=0.192]$ | | |
| 4th attempt: $(0.4)^2(0.6)^2 \times (3 \text{ or } ^3C_1)$ $[=0.1728]$ | | |
| 5th attempt: $(0.4)^2(0.6)^3 \times (4 \text{ or } ^4C_1)$ $[=0.13824]$ | | |
| $= 0.663,\ \dfrac{2072}{3125}$ | A1 | If either or both M marks not awarded, SC B1 for $0.663$, $\dfrac{2072}{3125}$ WWW condone 1 index error |

**Method 2**

| Answer | Mark | Guidance |
|--------|------|----------|
| $^5C_2(0.4)^2(0.6)^3 + ^5C_3(0.4)^3(0.6)^2 + ^5C_4(0.4)^4(0.6)^1 + ^5C_5(0.4)^5$ | M1 | At least 2 correct unsimplified terms |
| $[0.3456 + 0.2304 + 0.0768 + 0.01024]$ or $1-(^5C_0(0.6)^5 + ^5C_1(0.4)^1(0.6)^4)$ | M1 | Add values for 4 terms of the form $^5C_a(0.4)^a(0.6)^{5-a}$, or $1 -$ sum of 2 terms of the form $^5C_a(0.4)^a(0.6)^{5-a}$ |
| $=0.663,\ \dfrac{2072}{3125}$ | A1 | If either or both M marks not awarded, SC B1 for $0.663$ www condone 1 index error |

**Total: 3 marks**

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## Question 7(d):

| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{Mean} = 75 \times 0.4 =]\ 30$; $[\text{Variance} = 75 \times 0.4 \times 0.6 =]\ 18$ | B1 | 30 and 18 seen, allow unsimplified. $\sigma = \sqrt{18}, 3\sqrt{2}, 4.2426 \leq \sigma \leq 4.243$ implies correct variance. Withhold if variance clearly identified as standard deviation |
| $P(28 < X < 35) = P\!\left(\dfrac{28.5-30}{\sqrt{18}} < Z < \dfrac{34.5-30}{\sqrt{18}}\right)$ | M1 | Substituting *their* $\mu$ and positive $\sigma$ into one $\pm$ standardising formula (any number for 28.5 or 34.5), not $\sigma^2$, not $\sqrt{\sigma}$ |
| | M1 | Using continuity corrections $27.5$ *or* $28.5$ and $34.5$ *or* $35.5$ in *their* 2 separate standardisation formulae |
| $[= \Phi(1.0607) + \Phi(0.3536) - 1]$ $= 0.8556 + 0.6383 - 1$ or $0.8556-(1-0.6383)$ or $0.8556-0.3617$ or $(0.8556-0.5)+(0.6383-0.5)$ or $0.3556+0.1383$ | M1 | Appropriate area $\Phi$, from final process. Must be a probability |
| $= 0.494$ | A1 | AWRT |

**Total: 5 marks**