| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability of specific committee composition |
| Difficulty | Moderate -0.3 This is a standard S1 combinations question with straightforward application of nCr and probability rules. Part (i) is direct calculation, parts (ii)-(iii) require basic multiplication of combinations, and part (iv) needs complementary counting but follows textbook patterns. Slightly easier than average due to clear structure and routine methods. |
| Spec | 2.03a Mutually exclusive and independent events5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| \(x\) | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 10 }\) | \(\frac { 1 } { 5 }\) | \(\frac { 3 } { 10 }\) | \(\frac { 2 } { 5 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i)(a) | X-B(30, 0.05) seen or implied | B1 |
| 7(i)(b) | Addition method: X-B(30,0.05) & Y-B(15,0.05) stated or implied; P(X = 2) = (0.8122–0.5535) or \({}^{30}C_2×0.05^2×0.95^{28}\) or 0.2587/6; OR P(Y ≥ 1) = (1 – 0.95\(^{15}\)) or 0.5367 | M1 |
| fully correct method for P(X=2)×P(Y=0); "0.2587" × "0.5367" or 0.1388 M1 | M1 | [their (a)+any p] alone, but dep 1st MI |
| P(X > 2) + P(X = 2) × P(Y ≥ 1) = 0.1878" + "0.1388" alone | M1 | = 0.327 (3 sf) AG; For A1 must see correct wking or 0.326/56/...; For A0 must see correct wking out of 0.3265/6... |
| = 0.327 (3 sf) AG | A1 | If ans 0.327, check whether it comes from a correct method (possibly not in MS) or clearly comes from an incorrect method eg (0.4465 + 0.2587)×0.4633 = 0.327 B1M1M0M0A0 |
| 7(ii) | Any use of 0.327 or their (i)(b) for 1st M1; (1–0.327)³ ×0.327 + (1–0.327)⁴×0.327 Allow "correct" use of their (i)(a) or (i)(b) for 2nd M1 | M1, M1 |
**7(i)(a)** | X-B(30, 0.05) seen or implied | B1 | eg by 0.8122 seen; Bin(..., 0.6) B0B1. Can still score comment marks |
**7(i)(b)** | Addition method: X-B(30,0.05) & Y-B(15,0.05) stated or implied; P(X = 2) = (0.8122–0.5535) or ${}^{30}C_2×0.05^2×0.95^{28}$ or 0.2587/6; OR P(Y ≥ 1) = (1 – 0.95$^{15}$) or 0.5367 | M1 | NB eg 0.0362 implies B(15, 0.05) see below; P(X=2) = 0.8122–0.5535 or 0.2587/6 or 0.1388 M1 |
| fully correct method for P(X=2)×P(Y=0); "0.2587" × "0.5367" or 0.1388 M1 | M1 | [their (a)+any p] alone, but dep 1st MI |
| P(X > 2) + P(X = 2) × P(Y ≥ 1) = 0.1878" + "0.1388" alone | M1 | = 0.327 (3 sf) AG; For A1 must see correct wking or 0.326/56/...; For A0 must see correct wking out of 0.3265/6... |
| = 0.327 (3 sf) AG | A1 | If ans 0.327, check whether it comes from a correct method (possibly not in MS) or clearly comes from an incorrect method eg (0.4465 + 0.2587)×0.4633 = 0.327 B1M1M0M0A0 |
**7(ii)** | Any use of 0.327 or their (i)(b) for 1st M1; (1–0.327)³ ×0.327 + (1–0.327)⁴×0.327 Allow "correct" use of their (i)(a) or (i)(b) for 2nd M1 | M1, M1 | = 0.167 (3 sf) |7 A committee of 7 people is to be chosen at random from 18 volunteers.\\
(i) In how many different ways can the committee be chosen?
The 18 volunteers consist of 5 people from Gloucester, 6 from Hereford and 7 from Worcester. The committee is to be chosen randomly. Find the probability that the committee will\\
(ii) consist of 2 people from Gloucester, 2 people from Hereford and 3 people from Worcester,\\
(iii) include exactly 5 people from Worcester,\\
(iv) include at least 2 people from each of the three cities.
1 Jenny and John are each allowed two attempts to pass an examination.\\
(i) Jenny estimates that her chances of success are as follows.
\begin{itemize}
\item The probability that she will pass on her first attempt is $\frac { 2 } { 3 }$.
\item If she fails on her first attempt, the probability that she will pass on her second attempt is $\frac { 3 } { 4 }$. Calculate the probability that Jenny will pass.\\
(ii) John estimates that his chances of success are as follows.
\item The probability that he will pass on his first attempt is $\frac { 2 } { 3 }$.
\item Overall, the probability that he will pass is $\frac { 5 } { 6 }$.
\end{itemize}
Calculate the probability that if John fails on his first attempt, he will pass on his second attempt.
2 For each of 50 plants, the height, $h \mathrm {~cm}$, was measured and the value of ( $h - 100$ ) was recorded. The mean and standard deviation of $( h - 100 )$ were found to be 24.5 and 4.8 respectively.\\
(i) Write down the mean and standard deviation of $h$.
The mean and standard deviation of the heights of another 100 plants were found to be 123.0 cm and 5.1 cm respectively.\\
(ii) Describe briefly how the heights of the second group of plants compare with the first.\\
(iii) Calculate the mean height of all 150 plants.
3 In Mr Kendall's cupboard there are 3 tins of baked beans and 2 tins of pineapple. Unfortunately his daughter has removed all the labels for a school project and so the tins are identical in appearance. Mr Kendall wishes to use both tins of pineapple for a fruit salad. He opens tins at random until he has opened the two tins of pineapples.
Let $X$ be the number of tins that Mr Kendall opens.\\
(i) Show that $\mathrm { P } ( X = 3 ) = \frac { 1 } { 5 }$.\\
(ii) The probability distribution of $X$ is given in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 10 }$ & $\frac { 1 } { 5 }$ & $\frac { 3 } { 10 }$ & $\frac { 2 } { 5 }$ \\
\hline
\end{tabular}
\end{center}
Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR S1 Q7 [14]}}