| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Second success on trial n |
| Difficulty | Standard +0.3 Part (a) is a direct application of the negative binomial formula for second success. Part (b) requires standard assumptions (independence, constant probability). Part (c) involves using mean and variance formulas for negative binomial to find p, then comparing—straightforward algebra but requires knowing the formulas. Overall slightly easier than average as it's mostly formula application with minimal problem-solving. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\binom{7}{1} \times 0.15^2 \times (0.85)^6\) | M1 | Selecting appropriate model: negative binomial or \(B(7, 0.15)\) with success in 8th trial |
| \(= 0.05940\ldots =\) awrt \(\mathbf{0.0594}\) | A1 | awrt 0.0594 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The games (trials) are independent | B1 | Stating games are independent |
| The probability of winning a prize, 0.15, is constant for each game | B1 | Stating probability remains constant |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(18 = \frac{r}{p}\) and \(6^2 = \frac{r(1-p)}{p^2}\) | M1, A1 | M1: forming equation for mean or standard deviation. A1: both equations correct |
| Solving: \(2p = 1-p\) | M1 | Solving two equations leading to \(2p = 1-p\) |
| \(p = \frac{1}{3}\) (> 0.15) so Mary has the greater chance of winning a prize | A1 | For \(p = \frac{1}{3}\) followed by correct deduction |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\binom{7}{1} \times 0.15^2 \times (0.85)^6$ | M1 | Selecting appropriate model: negative binomial or $B(7, 0.15)$ with success in 8th trial |
| $= 0.05940\ldots =$ awrt $\mathbf{0.0594}$ | A1 | awrt 0.0594 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The games (trials) are **independent** | B1 | Stating games are independent |
| The probability of winning a prize, 0.15, is **constant** for each game | B1 | Stating probability remains constant |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $18 = \frac{r}{p}$ and $6^2 = \frac{r(1-p)}{p^2}$ | M1, A1 | M1: forming equation for mean or standard deviation. A1: both equations correct |
| Solving: $2p = 1-p$ | M1 | Solving two equations leading to $2p = 1-p$ |
| $p = \frac{1}{3}$ (> 0.15) so Mary has the greater chance of winning a prize | A1 | For $p = \frac{1}{3}$ followed by correct deduction |\begin{enumerate}
\item The probability of Richard winning a prize in a game at the fair is 0.15
\end{enumerate}
Richard plays a number of games.\\
(a) Find the probability of Richard winning his second prize on his 8th game,\\
(b) State two assumptions that have to be made, for the model used in part (a) to be valid.
M ary plays the same game, but has a different probability of winning a prize. She plays until she has won r prizes. The random variable $G$ represents the total number of games M ary plays.\\
(c) Given that the mean and standard deviation of G are 18 and 6 respectively, determine whether Richard or Mary has the greater probability of winning a prize in a game.\\
\hfill \mbox{\textit{Edexcel FS1 Q5 [8]}}