| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Probability distribution finding parameters |
| Difficulty | Moderate -0.8 This is a straightforward probability question requiring basic recall of fundamental principles. Part (i) uses the fact that probabilities sum to 1 (simple algebra: 0.3 + 0.15 + 3p + 2p + 0.05 = 1). Parts (ii)(a) and (ii)(b) involve listing outcomes and multiplying independent probabilities—routine calculations with no conceptual challenge or problem-solving insight required. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general |
| \(x\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( X = x )\) | 0.3 | 0.15 | \(3 p\) | \(2 p\) | 0.05 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(p = 0.1\) | B1 [1] | |
| (ii) (a) \(P(X = 1, Y = 3) = 0.3 \times 0.2 = 0.06\) | M1 | Summing 2 or 3 options |
| \(P(X = 2, Y = 2) = 0.15 \times 0.5 = 0.075\) | B1 | One option correct unsimplified |
| \(P(X = 3, Y = 1) = 0.3 \times 0.3 = 0.09\) | ||
| \(P(\text{sum is } 4) = 0.225\) | A1 [3] | correct final answer |
| (ii) (b) \(P(X = 1, Y = \text{anything}) = 0.3\) | M1 | \(\sum\) 3 or more two-factor options |
| \(P(X = 2, Y = \text{anything}) = 0.15\) | B1 | Two correct options |
| \(P(X = 3, Y = 1, 2) = 0.3 \times 0.8 = 0.24\) | ||
| \(P(X = 4, Y = 1) = 0.2 \times 0.3 = 0.06\) | ||
| \(P(X = 5, Y = 1) = 0.05 \times 0.3 = 0.015\) | ||
| \(P(\text{product} < 8) = 0.765\) | A1 [3] | Correct answer |
| OR \(P(Y = 1, X = \text{anything}) = 0.3\) | M1 | |
| \(P(Y = 2, X = 1, 2, 3) = 0.5 \times 0.75 = 0.375\) | B1 | |
| \(P(Y = 3, X = 1, 2) = 0.2 \times 0.45 = 0.09\) | ||
| \(P(\text{product} < 8) = 0.765\) | A1 |
**(i)** $p = 0.1$ | B1 [1]
**(ii) (a)** $P(X = 1, Y = 3) = 0.3 \times 0.2 = 0.06$ | M1 | Summing 2 or 3 options
$P(X = 2, Y = 2) = 0.15 \times 0.5 = 0.075$ | B1 | One option correct unsimplified
$P(X = 3, Y = 1) = 0.3 \times 0.3 = 0.09$ |
$P(\text{sum is } 4) = 0.225$ | A1 [3] | correct final answer
**(ii) (b)** $P(X = 1, Y = \text{anything}) = 0.3$ | M1 | $\sum$ 3 or more two-factor options
$P(X = 2, Y = \text{anything}) = 0.15$ | B1 | Two correct options
$P(X = 3, Y = 1, 2) = 0.3 \times 0.8 = 0.24$ |
$P(X = 4, Y = 1) = 0.2 \times 0.3 = 0.06$ |
$P(X = 5, Y = 1) = 0.05 \times 0.3 = 0.015$ |
$P(\text{product} < 8) = 0.765$ | A1 [3] | Correct answer
**OR** $P(Y = 1, X = \text{anything}) = 0.3$ | M1
$P(Y = 2, X = 1, 2, 3) = 0.5 \times 0.75 = 0.375$ | B1
$P(Y = 3, X = 1, 2) = 0.2 \times 0.45 = 0.09$ |
$P(\text{product} < 8) = 0.765$ | A13 A spinner has 5 sides, numbered 1, 2, 3, 4 and 5 . When the spinner is spun, the score is the number of the side on which it lands. The score is denoted by the random variable $X$, which has the probability distribution shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = x )$ & 0.3 & 0.15 & $3 p$ & $2 p$ & 0.05 \\
\hline
\end{tabular}
\end{center}
(i) Find the value of $p$.
A second spinner has 3 sides, numbered 1, 2 and 3. The score when this spinner is spun is denoted by the random variable $Y$. It is given that $\mathrm { P } ( Y = 1 ) = 0.3 , \mathrm { P } ( Y = 2 ) = 0.5$ and $\mathrm { P } ( Y = 3 ) = 0.2$.\\
(ii) Find the probability that, when both spinners are spun together,
\begin{enumerate}[label=(\alph*)]
\item the sum of the scores is 4,
\item the product of the scores is less than 8 .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2012 Q3 [7]}}