| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | r-th success on trial n |
| Difficulty | Standard +0.3 This is a straightforward application of geometric and negative binomial distributions with clear formulas. Part (a) tests standard probability calculations: (i) geometric distribution P(X=4), (ii) negative binomial for r-th success, (iii) binomial distribution. Part (b) requires computing cumulative probabilities but involves routine summation. All parts are direct formula applications with no conceptual tricks, making this slightly easier than average for Further Maths. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(X \sim \text{Geo}(0.2)\) or P\((X = 4) = 0.8^3 \times 0.2 = \mathbf{0.1024}\) | M1, A1 | 3.3, 1.1b |
| (ii) \(T \sim \text{NegBin}(3, 0.2)\) or P\((T = 8) = \binom{7}{2}0.2^2 \times 0.8^5 \times 0.2 = 0.05505\ldots\) awrt \(\mathbf{0.0551}\) | M1, A1 | 3.3, 1.1b |
| (iii) \(F \sim \text{B}(10, 0.2)\) or P\((F = 4) = \binom{10}{4}0.2^4 \times 0.8^6 = 0.088080\ldots\) awrt \(\mathbf{0.0881}\) | M1, A1 | 3.3, 1.1b |
| (b) \(\text{P}(R) = \text{P}(X \leq 4)\) and \(X \sim \text{Geo}(0.2)\) \(\mid\) P\((X \geq 1)\), \(X \sim \text{B}(4, 0.2)\) | M1, M1 | 3.1b, 3.4 |
| \(= 1 - \text{P}(X > 4) = 1 - 0.8^4\) \(\mid\) \(= 1 - \text{P}(Y = 0) = 1 - 0.8^4\) \(= \mathbf{0.59(04)}\) | A1, M1 | 1.1b, 3.1b |
| \(\text{P}(Y) = \text{P}(N \leq 7)\) and \(N \sim \text{NegBin}(3, 0.4)\) | M1 | |
| \(0.4^3\left(\begin{array}{l}2\end{array}\right)0.4^30.6^1 +\) | M1 | 3.4 |
| \(\binom{4}{2}0.4^30.6^2 + \binom{5}{2}0.4^30.6^3 +\left[\frac{7}{2}\right]0.4^30.6^6 + \binom{6}{2}0.4^30.6^4\) \(\left[\frac{1}{0.4^0\times 0.6^7}\right]\) | M1 | 3.4 |
| \(\text{P}(Y) = \text{P}(W > 2)\) where \(W \sim \text{B}(7, 0.4)\) | M1 | |
| \(= 1 - \text{P}(W \leq 2) [= 1 - 0.419904]\) \(= \mathbf{0.58(0096)}\) | A1 | 1.1b |
| ALT \(\text{P}(Y) = \text{P}(W > 2)\) where \(W \sim \text{B}(7, 0.4)\) | M1 | |
| \(= 1 - \text{P}(W \leq 2) [= 1 - 0.419904]\) \(= \mathbf{0.58(0096)}\) | A1 | 1.1b |
**(a)(i)** $X \sim \text{Geo}(0.2)$ or P$(X = 4) = 0.8^3 \times 0.2 = \mathbf{0.1024}$ | M1, A1 | 3.3, 1.1b | (2)
**(ii)** $T \sim \text{NegBin}(3, 0.2)$ or P$(T = 8) = \binom{7}{2}0.2^2 \times 0.8^5 \times 0.2 = 0.05505\ldots$ awrt $\mathbf{0.0551}$ | M1, A1 | 3.3, 1.1b | (2)
**(iii)** $F \sim \text{B}(10, 0.2)$ or P$(F = 4) = \binom{10}{4}0.2^4 \times 0.8^6 = 0.088080\ldots$ awrt $\mathbf{0.0881}$ | M1, A1 | 3.3, 1.1b | (2)
**(b)** $\text{P}(R) = \text{P}(X \leq 4)$ and $X \sim \text{Geo}(0.2)$ $\mid$ P$(X \geq 1)$, $X \sim \text{B}(4, 0.2)$ | M1, M1 | 3.1b, 3.4 |
$= 1 - \text{P}(X > 4) = 1 - 0.8^4$ $\mid$ $= 1 - \text{P}(Y = 0) = 1 - 0.8^4$ $= \mathbf{0.59(04)}$ | A1, M1 | 1.1b, 3.1b |
$\text{P}(Y) = \text{P}(N \leq 7)$ and $N \sim \text{NegBin}(3, 0.4)$ | M1 |
$0.4^3\left(\begin{array}{l}2\end{array}\right)0.4^30.6^1 +$ | M1 | 3.4 |
$\binom{4}{2}0.4^30.6^2 + \binom{5}{2}0.4^30.6^3 +\left[\frac{7}{2}\right]0.4^30.6^6 + \binom{6}{2}0.4^30.6^4$ $\left[\frac{1}{0.4^0\times 0.6^7}\right]$ | M1 | 3.4 |
$\text{P}(Y) = \text{P}(W > 2)$ where $W \sim \text{B}(7, 0.4)$ | M1 |
$= 1 - \text{P}(W \leq 2) [= 1 - 0.419904]$ $= \mathbf{0.58(0096)}$ | A1 | 1.1b | (7)
**ALT** $\text{P}(Y) = \text{P}(W > 2)$ where $W \sim \text{B}(7, 0.4)$ | M1 |
$= 1 - \text{P}(W \leq 2) [= 1 - 0.419904]$ $= \mathbf{0.58(0096)}$ | A1 | 1.1b | (7) | | (13)
**Notes:**
(a)(i) M1 for selecting the correct model. Stated or used which may be implied by ans.
A1 for 0.1024 or $\frac{64}{625}$ (accept 0.102) (correct answer scores 2 out of 2)
(ii) M1 for selecting the correct model. Stated or used which may be implied by ans. Allow $0.2 \times \text{P}(T = 2)$ from $T \sim \text{B}(7, 0.2)$
A1 for awrt 0.0551 (correct answer scores 2 out of 2)
(iii) M1 for selecting the correct model. Stated or used may be implied by ans of 0.967(2)
A1 for awrt 0.0881 (correct answer scores 2 out of 2)
(b) 1st M1 for a correct distribution and prob. expression for P$(R)$ (may be implied by 2nd M1)
2nd M1 for a correct numerical expression for P$(R)$ (allow any equivalent expression)
1st A1 for awrt 0.590 or $\frac{369}{625}$ (accept 0.59 or better) awrt 0.590 implies M1M1A1
3rd M1 for a correct distribution and prob. expression for P$(Y)$ (may be implied by 4th M1)
4th M1 for a correct numerical expression for P$(Y)$ (allow any equivalent expression)
2nd A1 for awrt 0.580 or (accept 0.58 or better) awrt 0.580 implies M1M1A1
**(Total: 13 marks)**\begin{enumerate}
\item Each time a spinner is spun, the probability that it lands on red is 0.2\\
(a) Find the probability that the spinner lands on red\\
(i) for the 1st time on the 4th spin\\
(ii) for the 3rd time on the 8th spin\\
(iii) exactly 4 times during 10 spins
\end{enumerate}
Each time the spinner is spun, the probability that it lands on yellow is 0.4\\
In a game with this spinner, a player must choose one of two events\\
$R$ is the event that the spinner lands on red for the $\mathbf { 1 s t }$ time in at most 4 spins\\
$Y$ is the event that the spinner lands on yellow for the 3rd time in at most 7 spins\\
(b) Showing your calculations clearly, determine which of these events has the greater probability.
\hfill \mbox{\textit{Edexcel FS1 2023 Q7 [13]}}