# Question 8:

## Part 8(a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X = 2) = 0.224$ | B1 | Correct probability AWRT 0.224 |

## Part 8(a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y \sim Po(18)$; $P(Y > 30) = P(Y \geq 31) = 0.0033$ | B1 | Selects Poisson model with $\lambda = 3 \times 6 = 18$ (PI); implied by 0.0059 for $P(Y \geq 30)$ |
| $P(Y \geq 31)$ or $1 - P(Y \leq 30)$ | M1 | Identifies correct probability |
| $0.0033$ (AWRT) | A1 | Correct probability |

## Part 8(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $C \sim B(6,\ 0.224)$ | B1 | Selects binomial model with $n=6$ and $p$ their probability from 8(a)(i) (PI) |
| $P(C=2) = \binom{6}{2}0.224^2(1-0.224)^4$ | M1 | Calculates $0.224^2(1-0.224)^4$ with their $0.224$ |
| $0.273$ (AWRT) | A1 | Correct probability |

## Part 8(c)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T \sim \text{Exp}(3)$; $P(T > 1) = e^{-3 \times 1}$ | B1 | Selects exponential or Poisson model with $\lambda = 3$ calls per 10 minutes or equivalent (PI) |
| $e^{-3}$ | M1 | Calculates correct probability consistent with their exponential or Poisson model; condone confusion between mean and parameter of exponential |
| $0.0498$ (AWRT) | A1 | Correct probability |

## Part 8(c)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.0498$ (follow through from 8(c)(i)) | B1F | States correct probability following through their answer to 8(c)(i) |
| Due to the memoryless property of the exponential distribution, probability is unaffected by the current time since the last call | E1 | Deduces it is the same because of the memoryless property of the exponential distribution |