| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success before/after trial n |
| Difficulty | Standard +0.3 Part (a) involves standard geometric and binomial distribution calculations that are routine for Further Statistics 1. Part (b) requires recognizing an infinite geometric series for alternating turns, which is a common Further Maths problem type but does require some problem-solving insight beyond pure recall. |
| Spec | 5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \([W \sim \text{Geo}(0.11)]\) \(P(W=6) = (0.89)^5(0.11)\) | M1 | Correct method to find \(P(W=6)\) e.g. \((p)^5(1-p)\) for \(p=0.11\) or \(0.89\) |
| \(= 0.06142\ldots\) awrt \(\mathbf{0.0614}\) | A1 | Correct ans with no incorrect working 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(W \text{ ,, } 5) = 1-(0.89)^5\) | M1 | Correct method to find \(P(W\text{ ,, }5)\) |
| \(= 0.44159\ldots\) awrt \(\mathbf{0.442}\) | A1 | Correct ans with no incorrect working 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim \text{B}(6, 0.11)\) | M1 | For using model B(6, 0.11); allow B(6, 0.89) [Implied by 0.0017 or awrt 0.114] |
| \(P(X=4) = 0.001739\ldots\) awrt \(\mathbf{0.00174}\) | A1 | Correct ans with no incorrect working 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| \([Y \sim \text{NB}(4, 0.11)]\) using neg bin or \(V \sim \text{B}(6, 0.11)\) and \(P(V\ldots 4)\) for M2 | M1 | For using a negative binomial model implied by correct \(P(Y=5)\) or \(P(Y=6)\) |
| \(P(Y\text{ ,, }6) = P(Y=4)+P(Y=5)+P(Y=6)\) | M1 | Correct method to find \(P(Y,, 6)\) |
| \(= (0.11)^4 + \binom{4}{3}(0.11)^3(0.89)^1\times0.11 + \binom{5}{3}(0.11)^3(0.89)^2\times0.11\) | M1 | At least two correct terms |
| \(= 0.001827\) awrt \(\mathbf{0.00183}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Zac wins}) = 0.89\times0.11 + (0.89)^3\times0.11 + (0.89)^5\times0.11+\ldots\) | M1 | Forming correct probability of Zac winning or identify \(a\) and \(r\) of GP. Allow for \(p=(0.11)\times0+(1-0.11)(1-p)\) |
| \(= \dfrac{0.89\times0.11}{1-(0.89)^2}\) | M1 | Using sum to infinity of GP. Allow for \(p=\dfrac{0.89}{1+0.89}\) |
| \(= 0.47089\ldots = 0.471^*\) | A1cso | Previous method marks must be seen leading to answer 0.471 (NOT awrt 0.471) |
## Question 4:
### Part (a)(i):
| $[W \sim \text{Geo}(0.11)]$ $P(W=6) = (0.89)^5(0.11)$ | M1 | Correct method to find $P(W=6)$ e.g. $(p)^5(1-p)$ for $p=0.11$ or $0.89$ |
| $= 0.06142\ldots$ awrt $\mathbf{0.0614}$ | A1 | Correct ans with no incorrect working 2/2 |
### Part (a)(ii):
| $P(W \text{ ,, } 5) = 1-(0.89)^5$ | M1 | Correct method to find $P(W\text{ ,, }5)$ |
| $= 0.44159\ldots$ awrt $\mathbf{0.442}$ | A1 | Correct ans with no incorrect working 2/2 |
### Part (a)(iii):
| $X \sim \text{B}(6, 0.11)$ | M1 | For using model B(6, 0.11); allow B(6, 0.89) [Implied by 0.0017 or awrt 0.114] |
| $P(X=4) = 0.001739\ldots$ awrt $\mathbf{0.00174}$ | A1 | Correct ans with no incorrect working 2/2 |
### Part (a)(iv):
| $[Y \sim \text{NB}(4, 0.11)]$ using neg bin **or** $V \sim \text{B}(6, 0.11)$ and $P(V\ldots 4)$ for M2 | M1 | For using a negative binomial model implied by correct $P(Y=5)$ or $P(Y=6)$ |
| $P(Y\text{ ,, }6) = P(Y=4)+P(Y=5)+P(Y=6)$ | M1 | Correct method to find $P(Y,, 6)$ |
| $= (0.11)^4 + \binom{4}{3}(0.11)^3(0.89)^1\times0.11 + \binom{5}{3}(0.11)^3(0.89)^2\times0.11$ | M1 | At least two correct terms |
| $= 0.001827$ awrt $\mathbf{0.00183}$ | A1 | |
### Part (b):
| $P(\text{Zac wins}) = 0.89\times0.11 + (0.89)^3\times0.11 + (0.89)^5\times0.11+\ldots$ | M1 | Forming correct probability of Zac winning or identify $a$ and $r$ of GP. Allow for $p=(0.11)\times0+(1-0.11)(1-p)$ |
| $= \dfrac{0.89\times0.11}{1-(0.89)^2}$ | M1 | Using sum to infinity of GP. Allow for $p=\dfrac{0.89}{1+0.89}$ |
| $= 0.47089\ldots = 0.471^*$ | A1cso | Previous method marks must be seen leading to answer 0.471 (NOT awrt 0.471) |
---\begin{enumerate}
\item In a game a spinner is spun repeatedly. When the spinner is spun, the probability of it landing on blue is 0.11\\
(a) Find the probability that the spinner lands on blue\\
(i) for the first time on the 6th spin,\\
(ii) for the first time before the 6th spin,\\
(iii) exactly 4 times during the first 6 spins,\\
(iv) for the 4th time on or before the 6th spin.
\end{enumerate}
Zac and Izana play the game. They take turns to spin the spinner. The winner is the first one to have the spinner land on blue. Izana spins the spinner first.\\
(b) Show that the probability of Zac winning is 0.471 to 3 significant figures.
\hfill \mbox{\textit{Edexcel FS1 2022 Q4 [13]}}