# Question 7:

## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[B \sim \text{Geo}(\tfrac{1}{3})]$ $P(B = 4) = \left(\tfrac{2}{3}\right)^3 \times \tfrac{1}{3}$ | M1 | For selecting correct model i.e. $\text{Geo}(p)$ (may be implied by correct expression) |
| $= \dfrac{8}{81}$ | A1 | For $\frac{8}{81}$ (= 0.098765... accept awrt 0.0988) |

## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(B \leq 5) = 1 - P(B > 5)$ or $1 - \left(\tfrac{2}{3}\right)^5$ | M1 | For suitable strategy to use geometric model to find correct expression |
| $= \dfrac{211}{243}$ | A1 | For $\frac{211}{243}$ (= 0.868312... accept awrt 0.868) |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(B^2) = \text{Var}(B) + [E(B)]^2$ | M1 | For suitable strategy to find $E(B^2)$ [allow $G''(1) + G'(1)$] |
| From formula booklet: $E(B) = \dfrac{1}{1/3} = 3$ and $\text{Var}(B) = \dfrac{1 - \frac{1}{3}}{(\frac{1}{3})^2} = 6$ | B1 | For use of correct formulae to find $E(B) = 3$ and $\text{Var}(B) = 6$ or $G''(1) = 12$ |
| So $E(B^2) = 6 + 9 = \mathbf{15}$ | A1 | For 15. **SC** Formula for $E(B^2)$: allow M1B1A0 for $E(B^2) = \frac{2-p}{p^2}$ (o.e.) |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[R = \text{no. of the spin when it first lands on red}]$ $X = R \sim \text{Geo}(\tfrac{2}{3})$ | M1 | For choosing suitable geometric model (sight of $\text{Geo}(\frac{2}{3})$ or at least 3 correct probabilities) |
| Require $E(e^X) = \displaystyle\sum_{x=1}^{\infty} e^x \left(\tfrac{1}{3}\right)^{x-1} \cdot \tfrac{2}{3}$ | M1 | For realising need for appropriate expected value and using $E(g(X))$ [need sum and $f(x)$]; NB simply finding $e^{E(X)} = e^{1.5} = \text{awrt } 4.48$ is M0 |
| $= \dfrac{2e}{3} \displaystyle\sum_{x=1}^{\infty} \left(\tfrac{e}{3}\right)^{x-1}$ | M1 | For suitable strategy to turn expression into a sum that can be found |
| $= \dfrac{2e}{3} \times \dfrac{1}{1 - \frac{e}{3}}$ or $\dfrac{2e}{3-e}$ | A1 | For correct use of sum to infinity of geometric series |
| $E(e^X) = 19.297...\ \{> 15 = E(B^2)\}$ so Tamara should **choose red** since it has the greater expected score | A1 | For interpreting outcome in terms of solution to problem; must choose red and see awrt 19.3 (and allow ft of their $E(B^2) < 19$) |

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