| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Multi-stage game or match outcomes |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question testing basic probability tree concepts. Part (i) requires simple calculation using the complement rule and addition of mutually exclusive events. Part (ii) involves basic conditional probability formula P(A|B) = P(A∩B)/P(B), which is standard S1 material requiring minimal problem-solving beyond direct application of formulas. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3}{4} + \frac{1}{4} × \frac{3}{8} + \frac{1}{4} × \frac{5}{8} × \frac{3}{16}\) = \(\frac{447}{512}\) or 0.873 (3 sf) | M1, M1, A1, [3] | \(\frac{1}{4} × \frac{3}{8} × \frac{13}{16}\) (= \(\frac{65}{512}\) or 0.127). \(1 - \frac{1}{4} × \frac{5}{8} × \frac{13}{16}\) |
| Answer | Marks | Guidance |
|---|---|---|
| 0.6p or equiv seen; 0.4 + 0.6p = 0.58; p = 0.3 | B1, M1, A1, [3] | Tree diag alone insufficient for mark. Or 0.6p = 0.18. "0.18" alone insufficient. NB 0.6 × 0.3 = 0.18 seen at the end is probably a check, not an answer. But if 0.3 seen and 0.18 is very clearly indicated as the ans then B1M1A0 |
## (i)
$\frac{3}{4} + \frac{1}{4} × \frac{3}{8} + \frac{1}{4} × \frac{5}{8} × \frac{3}{16}$ = $\frac{447}{512}$ or 0.873 (3 sf) | M1, M1, A1, [3] | $\frac{1}{4} × \frac{3}{8} × \frac{13}{16}$ (= $\frac{65}{512}$ or 0.127). $1 - \frac{1}{4} × \frac{5}{8} × \frac{13}{16}$
## (ii)
0.6p or equiv seen; 0.4 + 0.6p = 0.58; p = 0.3 | B1, M1, A1, [3] | Tree diag alone insufficient for mark. Or 0.6p = 0.18. "0.18" alone insufficient. NB 0.6 × 0.3 = 0.18 seen at the end is probably a check, not an answer. But if 0.3 seen and 0.18 is very clearly indicated as the ans then B1M1A0 | NB 0.6 × 0.3 = 0.18 seen at the end is probably a check, not an answer. But if 0.3 seen and 0.18 is very clearly indicated as the ans then B1M1A0
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\item Kathryn is allowed three attempts at a high jump. If she succeeds on any attempt, she does not jump again. The probability that she succeeds on her first attempt is $\frac{1}{4}$. If she fails on her first attempt, the probability that she succeeds on her second attempt is $\frac{1}{3}$. If she fails on her first two attempts, the probability that she succeeds on her third attempt is $\frac{1}{2}$. Find the probability that she succeeds. [3]
\item Khaled is allowed two attempts to pass an examination. If he succeeds on his first attempt, he does not make a second attempt. The probability that he passes at the first attempt is 0.4 and the probability that he passes on either the first or second attempt is 0.58. Find the probability that he passes on the second attempt, given that he failed on the first attempt. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q2 [6]}}