| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Moderate -0.3 This is a straightforward application of standard geometric and binomial distribution formulas. Parts (i)-(iii) require direct use of geometric distribution probability, mean, and variance formulas with no conceptual challenges. Parts (iv)-(v) involve routine binomial probability calculations. All parts are textbook-standard with clear setups and no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.02d Binomial: mean np and variance np(1-p)5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) | 0.885 × 0.12 |
| = 0.0633 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | For (1 – p)5×p with their p |
| cao | Accept 0.06333 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | (0.886 = 0.46440…. = ) 0.464 to 3sf | B1 |
| [1] | 1.1 | Accept 0.4644 |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) | A | (1/ = ) 8.33 to 3 sf |
| 0.12 | B1 | |
| [1] | 1.1 | For 25/3 oe |
| Answer | Marks | Guidance |
|---|---|---|
| B | = 61.1 | B1 |
| [1] | 1.1 | For 61.1 or better |
| (iv) | B(10, 0.12) stated or 1 – 0.658… seen | |
| 0.342 (0.34172...) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | Accept 0.34 www |
| Answer | Marks |
|---|---|
| (v) | 9 × 0.888 × 0.122 |
| = 0.0466 (0.046608…) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1a | |
| 1.1 | 9 × (1 – p)8 × p2 with their p | Accept 0.047 www |
Question 4:
4 | (i) | 0.885 × 0.12
= 0.0633 | M1
A1
[2] | 3.3
1.1 | For (1 – p)5×p with their p
cao | Accept 0.06333
Accept 0.063 www
(ii) | (0.886 = 0.46440…. = ) 0.464 to 3sf | B1
[1] | 1.1 | Accept 0.4644
Accept 0.46 www
(iii) | A | (1/ = ) 8.33 to 3 sf
0.12 | B1
[1] | 1.1 | For 25/3 oe | Accept 8.3www
Accept 8.333 or better
B | = 61.1 | B1
[1] | 1.1 | For 61.1 or better
(iv) | B(10, 0.12) stated or 1 – 0.658… seen
0.342 (0.34172...) | M1
A1
[2] | 3.3
1.1 | Accept 0.34 www
Accept 0.3417
(v) | 9 × 0.888 × 0.122
= 0.0466 (0.046608…) | M1
A1
[2] | 1.1a
1.1 | 9 × (1 – p)8 × p2 with their p | Accept 0.047 www
Accept 0.046614 The probability that an expert darts player hits the bullseye on any throw is 0.12 , independently of any other throw. The player throws darts at the bullseye until she hits it.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the player has to throw exactly six darts.
\item Find the probability that the player has to throw more than six darts.
\item (A) Find the mean number of darts that the player has to throw.\\
(B) Find the variance of the number of darts that the player has to throw.
The player continues to throw more darts at the bullseye after she has hit it for the first time.
\item Find the probability that the player hits the bullseye at least twice in the first ten throws.
\item Find the probability that the player hits the bullseye for the second time on the tenth throw.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2018 Q4 [9]}}