| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Multiple independent binomial calculations |
| Difficulty | Standard +0.3 This question involves standard binomial distribution calculations with clearly stated probabilities. Parts (i) and (iii) are direct applications of binomial probability formulas, while part (ii) requires multinomial probability (which can be solved using combinations and independent probabilities). The normal approximation in part (iii) is a routine technique taught in S1. No novel insight or complex problem-solving is required—just careful application of standard methods. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(2) = {}^7C_2(0.1)^2(0.9)^5 = 0.124\) | M1, A1 [2] | Bin term \({}^7C_2 p^2(1-p)^5\), \(0 < p < 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((0.15)^1(0.1)^2(0.75)^2 \times \dfrac{5!}{1!2!2!}\) | M1 | Mult probs for options \((0.15)^a(0.1)^b(0.75)^c\) where \(a+b+c\) sum to 5 |
| \(= 0.0253\) or \(81/3200\) | M1, A1 [3] | Mult by \(\dfrac{5!}{1!2!2!}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| mean \(= 365\times0.15\ (= 54.75\) or \(219/4)\) | B1 | Correct unsimplified mean and var, oe |
| Var \(= 365\times0.15\times0.85\ (= 46.5375\) or \(3723/80)\) | ||
| \(P(x > 44) = P\!\left(z > \dfrac{44.5 - 54.75}{\sqrt{46.5375}}\right)\) | M1 | \(\pm\) standardising, need sq rt |
| M1 | cc either 44.5 (or 43.5) | |
| \(= P(z > -1.5025)\) | M1 | \(\Phi\) |
| \(= 0.933\) | A1 [5] | Correct answer; accept 0.934 |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(2) = {}^7C_2(0.1)^2(0.9)^5 = 0.124$ | M1, A1 [2] | Bin term ${}^7C_2 p^2(1-p)^5$, $0 < p < 1$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0.15)^1(0.1)^2(0.75)^2 \times \dfrac{5!}{1!2!2!}$ | M1 | Mult probs for options $(0.15)^a(0.1)^b(0.75)^c$ where $a+b+c$ sum to 5 |
| $= 0.0253$ or $81/3200$ | M1, A1 [3] | Mult by $\dfrac{5!}{1!2!2!}$ oe |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| mean $= 365\times0.15\ (= 54.75$ or $219/4)$ | B1 | Correct unsimplified mean **and** var, oe |
| Var $= 365\times0.15\times0.85\ (= 46.5375$ or $3723/80)$ | | |
| $P(x > 44) = P\!\left(z > \dfrac{44.5 - 54.75}{\sqrt{46.5375}}\right)$ | M1 | $\pm$ standardising, need sq rt |
| | M1 | cc either 44.5 (or 43.5) |
| $= P(z > -1.5025)$ | M1 | $\Phi$ |
| $= 0.933$ | A1 [5] | Correct answer; accept 0.934 |7 Each day Annabel eats rice, potato or pasta. Independently of each other, the probability that she eats rice is 0.75 , the probability that she eats potato is 0.15 and the probability that she eats pasta is 0.1 .\\
(i) Find the probability that, in any week of 7 days, Annabel eats pasta on exactly 2 days.\\
(ii) Find the probability that, in a period of 5 days, Annabel eats rice on 2 days, potato on 1 day and pasta on 2 days.\\
(iii) Find the probability that Annabel eats potato on more than 44 days in a year of 365 days.
\hfill \mbox{\textit{CAIE S1 2016 Q7 [10]}}