| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.8 This question combines Poisson distribution with geometric/negative binomial concepts. Part (a) is straightforward Poisson (P(X≥3) with λ=0.8), but parts (b) and (c) require recognizing the negative binomial distribution for waiting until the 3rd success, which is non-standard for FS1 and requires careful probability reasoning beyond routine application. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02g 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(X\) = no. of prizes Andreia wins, \(X \sim B(40, 0.02)\) | M1 | For selecting a suitable model i.e. \(B(40, p)\) where \(p\) is any probability. Written or used, may be implied by a correct answer or \(0.037429\ldots\) from \(P(X=3)\) |
| \(P(X \geqslant 3) = 1 - P(X \leqslant 2) = 0.04567\ldots\) awrt \(\mathbf{0.0457}\) | A1 | For awrt \(0.0457\) (correct answer only 2/2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(Y\) = no. of the bar when Barney wins, \(Y \sim \text{NegBin}(3, 0.02)\) | M1 | 1st M1 for selecting a suitable model \(NB(3, 0.02)\). May be implied by a correct expression |
| \(P(Y=40) = \binom{39}{2} \times 0.02^2 \times 0.98^{37} \times 0.02\) | M1 | 2nd M1 for use of model to form a correct expression. SC: \(p \neq 0.02\): allow prob of form \(\binom{39}{2}p^3(1-p)^{37}\) where \(0 < p < 1\) scores M0M1 |
| \(= 0.0028071\ldots\) awrt \(\mathbf{0.00281}\) | A1 | For awrt \(0.00281\) (accept awrt \(2.81 \times 10^{-3}\)) [correct answer with no working scores 3/3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y) = \dfrac{3}{0.02} = \mathbf{150}\) | B1 | B1 for 150 |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $X$ = no. of prizes Andreia wins, $X \sim B(40, 0.02)$ | M1 | For selecting a suitable model i.e. $B(40, p)$ where $p$ is any probability. Written or used, may be implied by a correct answer or $0.037429\ldots$ from $P(X=3)$ |
| $P(X \geqslant 3) = 1 - P(X \leqslant 2) = 0.04567\ldots$ awrt $\mathbf{0.0457}$ | A1 | For awrt $0.0457$ (correct answer only 2/2) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $Y$ = no. of the bar when Barney wins, $Y \sim \text{NegBin}(3, 0.02)$ | M1 | 1st M1 for selecting a suitable model $NB(3, 0.02)$. May be implied by a correct expression |
| $P(Y=40) = \binom{39}{2} \times 0.02^2 \times 0.98^{37} \times 0.02$ | M1 | 2nd M1 for use of model to form a correct expression. SC: $p \neq 0.02$: allow prob of form $\binom{39}{2}p^3(1-p)^{37}$ where $0 < p < 1$ scores M0M1 |
| $= 0.0028071\ldots$ awrt $\mathbf{0.00281}$ | A1 | For awrt $0.00281$ (accept awrt $2.81 \times 10^{-3}$) [correct answer with no working scores 3/3] |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y) = \dfrac{3}{0.02} = \mathbf{150}$ | B1 | B1 for 150 |
---\begin{enumerate}
\item A chocolate manufacturer places special tokens in $2 \%$ of the bars it produces so that each bar contains at most one token. Anyone who collects 3 of these tokens can claim a prize.
\end{enumerate}
Andreia buys a box of 40 bars of the chocolate.\\
(a) Find the probability that Andreia can claim a prize.
Barney intends to buy bars of the chocolate, one at a time, until he can claim a prize.\\
(b) Find the probability that Barney can claim a prize when he buys his 40th bar of chocolate.\\
(c) Find the expected number of bars that Barney must buy to claim a prize.
\hfill \mbox{\textit{Edexcel FS1 2019 Q1 [6]}}