# Question 2:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 'Randomly' means that **mutations** occur with no (predictable) pattern | E1 | In context. Allow "not predictable" |
| 'Independently' means that the occurrence of one **mutation** does not affect the **probability** of another mutation occurring | E1 | Must include 'probability' and context. Allow "chance". If not indicated, assume first comment relates to randomness. |
| **[2]** | | |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Exactly one}) = \binom{30}{29} \times 0.85^{29} \times 0.15^1 = {}^{20}C_1 \times 0.012^1 \times 0.988^{19}$ | M1 | For correct structure i.e. $20p(1-p)^{19}$ |
| $= 0.1908$ | A1 | Allow 0.191. Allow 0.19 www. |
| **[2]** | | |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Because the number of mutating genes $X$ is binomially distributed | E1 | Allow B(500, 0.012) or B$(n,p)$ |
| $n$ is large and $p$ is small | E1 | Allow the sample is large & $np \approx np(1-p)$ or $np$ not too large. Condone suitable numerical ranges e.g. $n > 30$, $p < 0.1$. Do not allow "the number is large and probability is small". Allow "probability of success/a gene mutating is small" for $p$ is small |
| **[2]** | | |

## Part (iv)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 500 \times 0.012 = 6$ | B1 | For mean |
| $P(2 \text{ mutations}) = e^{-0.85}\dfrac{0.85^2}{1!}$ $[0.0620 - 0.0174 \text{ from tables}]$ | M1 | Correct structure for $P(=2)$ using Poisson pdf or tables. |
| $= 0.0446$ | A1 | CAO Allow 0.04462 or 0.045www |
| **[3]** | | |

## Part (iv)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| From tables $P(\text{At least two}) = 1 - P(\leq 1)$ $= 1 - 0.0174 = 0.9826$ | M1 | For using $1 - P(\leq 1)$ using their mean. |
| | A1 | CAO Allow 0.983. Allow 0.98 www. |
| **[2]** | | |

## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $50000 \times 0.012 = 600$, Var $= 50000 \times 0.012 \times 0.988 = 592.8$ | B1 | For Normal approximation (SOI). |
| Using Normal approx. to the binomial, $X \sim N(600, 592.8)$ | B1 | For correct parameters (SOI). |
| $P(X \geq 650) = P\!\left(Z \leq \dfrac{30.5 - 25.5}{\sqrt{25.5}}\right)$ | B1 | For 649.5 |
| $= P(Z > 2.033) = 1 - \Phi(2.033) = 1 - 0.9789$ | M1 | For standardisation and probability calculation using correct tail. |
| $= 0.0211$ | A1 | CAO (Allow answer from calculator 0.0210) |
| **[5]** | | |

**Alternative solution using Normal approx. to Poisson:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $100 \times 6 = 600$, $X \sim N(600, 600)$ | B1, B1 | For Normal approximation (SOI). For correct parameters (SOI). |
| $P(X \geq 650) = P\!\left(Z \leq \dfrac{30.5-25.5}{\sqrt{25.5}}\right)$ | B1 | For 649.5 |
| $= P(Z > 2.021) = 1 - \Phi(2.021) = 1 - 0.9783$ | M1 | For standardisation and probability calculation using correct tail. |
| $= 0.0217$ | A1 | CAO (Allow answer from calculator 0.0216) |
| **[5]** | | |

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