| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Geometric distribution (first success) |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distribution and geometric distribution with standard calculations. Part (i) requires summing binomial probabilities P(X=8)+P(X=9)+P(X=10), part (ii) is a direct geometric distribution calculation (0.35)³(0.65), and part (iii) involves normal approximation to binomial. All are routine techniques for S1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \({}^{12}C_8(0.65)^8(0.35)^4 + {}^{12}C_9(0.65)^9(0.35)^3 + {}^{12}C_{10}(0.65)^{10}(0.35)^2\) | M1 | Bin term with \({}^{12}C_r p^r(1-p)^{12-r}\) seen, \(r \neq 0\), any \(p < 1\) |
| M1 | Summing 2 or 3 bin probs, \(p = 0.65\) or \(0.35\), \(n = 12\) | |
| \(= 0.541\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\overline{R}\,\overline{R}\,\overline{R}\,R) = 0.35 \times 0.35 \times 0.35 \times 0.65\) | M1 | Mult 4 probs either \((0.35)^3(0.65)\) or \((0.65)^3(0.35)\) |
| \(= 0.0279\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(7) = 0.2039\) (unsimplified) | B1 | \({}^{12}C_7(0.65)^7(0.35)^5\) |
| Mean \(= 250 \times 0.2039'\ (= 50.9798)\) | B1 | Correct unsimplified \(np\) and \(npq\) using 'their \(0.2039\)' but not \(0.65\) or \(0.35\) |
| Var \(= 250 \times 0.2039' \times (1 - 0.2039)'\ (= 40.5851)\) | ||
| \(P(>54) = P\!\left(\dfrac{54.5 - 50.9798}{\sqrt{40.5851}}\right)\) | M1 | Standardising – need \(\sqrt{\phantom{x}}\); must be from working with 54 |
| \(= P(z > 0.5526)\) | M1 | cc either 53.5 or 54.5 |
| \(= 1 - \Phi(0.5526) = 1 - 0.7098\) | M1 | Correct area \(< 0.5\), i.e. \(1 - \Phi\); must be from working with 54 |
| \(= 0.290\) | A1 [6] |
## Question 7:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| ${}^{12}C_8(0.65)^8(0.35)^4 + {}^{12}C_9(0.65)^9(0.35)^3 + {}^{12}C_{10}(0.65)^{10}(0.35)^2$ | **M1** | Bin term with ${}^{12}C_r p^r(1-p)^{12-r}$ seen, $r \neq 0$, any $p < 1$ |
| | **M1** | Summing 2 or 3 bin probs, $p = 0.65$ or $0.35$, $n = 12$ |
| $= 0.541$ | **A1** [3] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\overline{R}\,\overline{R}\,\overline{R}\,R) = 0.35 \times 0.35 \times 0.35 \times 0.65$ | **M1** | Mult 4 probs either $(0.35)^3(0.65)$ or $(0.65)^3(0.35)$ |
| $= 0.0279$ | **A1** [2] | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(7) = 0.2039$ (unsimplified) | **B1** | ${}^{12}C_7(0.65)^7(0.35)^5$ |
| Mean $= 250 \times 0.2039'\ (= 50.9798)$ | **B1** | Correct unsimplified $np$ and $npq$ using 'their $0.2039$' but not $0.65$ or $0.35$ |
| Var $= 250 \times 0.2039' \times (1 - 0.2039)'\ (= 40.5851)$ | | |
| $P(>54) = P\!\left(\dfrac{54.5 - 50.9798}{\sqrt{40.5851}}\right)$ | **M1** | Standardising – need $\sqrt{\phantom{x}}$; must be from working with 54 |
| $= P(z > 0.5526)$ | **M1** | cc either 53.5 or 54.5 |
| $= 1 - \Phi(0.5526) = 1 - 0.7098$ | **M1** | Correct area $< 0.5$, i.e. $1 - \Phi$; must be from working with 54 |
| $= 0.290$ | **A1** [6] | |7 Passengers are travelling to Picton by minibus. The probability that each passenger carries a backpack is 0.65 , independently of other passengers. Each minibus has seats for 12 passengers.\\
(i) Find the probability that, in a full minibus travelling to Picton, between 8 passengers and 10 passengers inclusive carry a backpack.\\
(ii) Passengers get on to an empty minibus. Find the probability that the fourth passenger who gets on to the minibus will be the first to be carrying a backpack.\\
(iii) Find the probability that, of a random sample of 250 full minibuses travelling to Picton, more than 54 will contain exactly 7 passengers carrying backpacks.
\hfill \mbox{\textit{CAIE S1 2016 Q7 [11]}}