| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Independent binomial samples with compound probability |
| Difficulty | Standard +0.3 This is a standard two-stage binomial problem: part (i) requires direct application of binomial probability formulas with n=10, p=0.25, while part (ii) involves using the complement rule with the result from (i)b as the probability for a second binomial with n=6. The question is slightly above average difficulty due to the compound probability structure in part (ii), but follows a well-established pattern taught in S1 with no novel insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Binomial seen or implied | B1 | by tables or \(^{10}C_3\) or \(^{10}C_7\), or by \(0.25^a \times 0.75^b\) \((a+b=10)\) |
| \(0.7759 - 0.5256\) or \(^{10}C_3 \times (1-0.25)^7 \times 0.25^3\) | M1 | Allow 0.25 |
| \(= 0.250\) (3 sf) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - 0.5256\) or \(1-((1-0.25)^{10} + 10(1-0.25)^9 \times 0.25 + {^{10}C_2}(1-0.25)^8 \times 0.25^2)\) | M1 | or \(P(X=3,4,5,6,7,8,9,10)\) all correct terms; Allow \(^{10}C_8\) instead of \(^{10}C_2\); NOT \(1-0.7759\) \((P(X>3))\) from table |
| \(= 0.4744\) or \(0.474\) (3 sf) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4744\) or \(0.474\) or \(0.5256\) or \(0.526\) seen | M1 | Their (i)(b) seen, or result of 1-(i)(b) seen; e.g. \(B(6, 0.474)\) or \(P(X \geq 3) = 0.474\) |
| \(1-(1-\text{"0.4744"})^6\) oe | M1 | or \(P(X=1,2,3,4,5,6)\) all correct terms seen |
| \(= 0.979\) (3 sf) | A1f | ft from (i)(b) |
| [3] |
# Question 7:
## Part (i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Binomial seen or implied | B1 | by tables or $^{10}C_3$ or $^{10}C_7$, or by $0.25^a \times 0.75^b$ $(a+b=10)$ |
| $0.7759 - 0.5256$ or $^{10}C_3 \times (1-0.25)^7 \times 0.25^3$ | M1 | Allow 0.25 |
| $= 0.250$ (3 sf) | A1 | |
| **[3]** | | |
## Part (i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 0.5256$ or $1-((1-0.25)^{10} + 10(1-0.25)^9 \times 0.25 + {^{10}C_2}(1-0.25)^8 \times 0.25^2)$ | M1 | or $P(X=3,4,5,6,7,8,9,10)$ all correct terms; Allow $^{10}C_8$ instead of $^{10}C_2$; NOT $1-0.7759$ $(P(X>3))$ from table |
| $= 0.4744$ or $0.474$ (3 sf) | A1 | |
| **[2]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4744$ or $0.474$ or $0.5256$ or $0.526$ seen | M1 | Their (i)(b) seen, or result of 1-(i)(b) seen; e.g. $B(6, 0.474)$ or $P(X \geq 3) = 0.474$ |
| $1-(1-\text{"0.4744"})^6$ oe | M1 | or $P(X=1,2,3,4,5,6)$ all correct terms seen |
| $= 0.979$ (3 sf) | A1f | ft from **(i)(b)** |
| **[3]** | | |
---7 Froox sweets are packed into tubes of 10 sweets, chosen at random. $25 \%$ of Froox sweets are yellow.\\
(i) Find the probability that in a randomly selected tube of Froox sweets there are
\begin{enumerate}[label=(\alph*)]
\item exactly 3 yellow sweets,
\item at least 3 yellow sweets.\\
(ii) Find the probability that in a box containing 6 tubes of Froox sweets, there is at least 1 tube that contains at least 3 yellow sweets.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2015 Q7 [8]}}