| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on constructing and using probability distributions. Part (i) involves basic probability setup with constraints, part (ii) is a guided 'show that' calculation, part (iii) uses the standard variance formula, and parts (iv)-(v) apply basic conditional probability. All steps are routine applications of S1 formulas with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.03d Calculate conditional probability: from first principles2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 4 | 6 | 7 | 8 | 10 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 3 } { 9 }\) | \(\frac { 1 } { 9 }\) | \(\frac { 2 } { 9 }\) | \(\frac { 2 } { 9 }\) | \(\frac { 1 } { 9 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) If \(y = P(\text{odd number})\) then \(P(\text{even number}) = 2y\) | M1 | 2P(Odd) shown = P(Even) and summed to 1 |
| \(3y + 6y = 1\) so \(y = 1/9\) oe. OR prob = 1/3 | A1 | correct answer accept either |
| [2] | ||
| (ii) Score of 8 means throwing a 6 | B1 | |
| \(6 \text{ is even so } P(8) = 2/9\) (AG) | B1 | legit justification of use of 2/9 |
| [2] | ||
| (iii) \(\text{Var}(X) = (48 + 36 + 98 + 128 + 100)/9 - (58/9)^2\) | M1 | Correct method no dividings, 6.44 squared subt numerically |
| \(= 4.02 \text{ accept } 4.025 \text{ (326/81)}\) | A1 | Correct answer |
| [2] | ||
| (iv) \(P(\text{score } 6,10) + P(\text{score } 10,6) + P(\text{score } 8,8)\) | M1 | Summing two different 2-factor probabilities |
| \(= 1/81 + 1/81 + 4/81 = 6/81 \text{ (2/27) (0.0741)}\) | A1 | Correct answer |
| [2] | ||
| (v) \(P(\text{score } 6, 10) = 1/81\) | B1 | 1/81 seen in numerator |
| \(P(1^{\text{st}} \text{ score } 6 \text{ given total } 16) = (1/81) \div (6/81) = 1/6\) | M1 | Dividing by their (iv) |
| A1 | Correct answer | |
| [3] |
**(i)** If $y = P(\text{odd number})$ then $P(\text{even number}) = 2y$ | M1 | 2P(Odd) shown = P(Even) and summed to 1
$3y + 6y = 1$ so $y = 1/9$ oe. OR prob = 1/3 | A1 | correct answer accept either
| [2] |
**(ii)** Score of 8 means throwing a 6 | B1 |
$6 \text{ is even so } P(8) = 2/9$ (AG) | B1 | legit justification of use of 2/9
| [2] |
**(iii)** $\text{Var}(X) = (48 + 36 + 98 + 128 + 100)/9 - (58/9)^2$ | M1 | Correct method no dividings, 6.44 squared subt numerically
$= 4.02 \text{ accept } 4.025 \text{ (326/81)}$ | A1 | Correct answer
| [2] |
**(iv)** $P(\text{score } 6,10) + P(\text{score } 10,6) + P(\text{score } 8,8)$ | M1 | Summing two different 2-factor probabilities
$= 1/81 + 1/81 + 4/81 = 6/81 \text{ (2/27) (0.0741)}$ | A1 | Correct answer
| [2] |
**(v)** $P(\text{score } 6, 10) = 1/81$ | B1 | 1/81 seen in numerator
$P(1^{\text{st}} \text{ score } 6 \text{ given total } 16) = (1/81) \div (6/81) = 1/6$ | M1 | Dividing by their (iv)
| A1 | Correct answer
| [3] |7 Sanket plays a game using a biased die which is twice as likely to land on an even number as on an odd number. The probabilities for the three even numbers are all equal and the probabilities for the three odd numbers are all equal.\\
(i) Find the probability of throwing an odd number with this die.
Sanket throws the die once and calculates his score by the following method.
\begin{itemize}
\item If the number thrown is 3 or less he multiplies the number thrown by 3 and adds 1 .
\item If the number thrown is more than 3 he multiplies the number thrown by 2 and subtracts 4 .
\end{itemize}
The random variable $X$ is Sanket's score.\\
(ii) Show that $\mathrm { P } ( X = 8 ) = \frac { 2 } { 9 }$.
The table shows the probability distribution of $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 4 & 6 & 7 & 8 & 10 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 3 } { 9 }$ & $\frac { 1 } { 9 }$ & $\frac { 2 } { 9 }$ & $\frac { 2 } { 9 }$ & $\frac { 1 } { 9 }$ \\
\hline
\end{tabular}
\end{center}
(iii) Given that $\mathrm { E } ( X ) = \frac { 58 } { 9 }$, find $\operatorname { Var } ( X )$.
Sanket throws the die twice.\\
(iv) Find the probability that the total of the scores on the two throws is 16 .\\
(v) Given that the total of the scores on the two throws is 16 , find the probability that the score on the first throw was 6 .
\hfill \mbox{\textit{CAIE S1 2010 Q7 [11]}}