| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | At least one success in repeated trials |
| Difficulty | Standard +0.3 This is a standard conditional probability problem requiring careful application of probability rules and tree diagram logic. Part (a) involves setting up equations using given probabilities and the complement rule (5 marks suggests multi-step working), while part (b) is a straightforward conditional probability calculation once part (a) is complete. The question requires systematic reasoning but uses familiar S1 techniques without novel insight. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Answer | Marks |
|---|---|
| (a) Let \(P(\text{miss after hit}) = x\) | \(M1\) \(M1\) |
| \(0.75 + 0.25x = 0.9\) | |
| \(x = 0.6\) | \(A1\) \(M1\) \(A1\) |
| \(P(H, M) = 0.25 \times 0.6 = 0.15\) | |
| (b) \(P[(M, M) \text{ at least 1 miss}] = (0.75 \times 0.7) + 0.9 = 0.583\) | \(M1\) \(A1\) \(A1\) |
**(a)** Let $P(\text{miss after hit}) = x$ | $M1$ $M1$ |
$0.75 + 0.25x = 0.9$ | |
$x = 0.6$ | $A1$ $M1$ $A1$ |
$P(H, M) = 0.25 \times 0.6 = 0.15$ | |
**(b)** $P[(M, M) \text{ at least 1 miss}] = (0.75 \times 0.7) + 0.9 = 0.583$ | $M1$ $A1$ $A1$ |
**Total: 8 marks**A darts player throws two darts, attempting to score a bull's-eye with each. The probability that he will achieve this with his first dart is $0.25$. If he misses with his first dart, the probability that he will also miss with his second dart is $0.7$. The probability that he will miss with at least one dart is $0.9$.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that he succeeds with his first dart but misses with his second is $0.15$. [5 marks]
\item Find the conditional probability that he misses with both darts, given that he misses with at least one. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q2 [8]}}