| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Selection with replacement |
| Difficulty | Easy -1.3 This is a straightforward S1 probability question testing basic concepts: independent events with replacement (parts a-b) and combinations without replacement (parts c-d). All parts use standard formulas with simple arithmetic—no problem-solving insight required, just routine application of probability rules taught early in the statistics module. |
| Spec | 2.03a Mutually exclusive and independent events5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{both blue}) = \frac{1}{20} \times \frac{1}{20} = \frac{1}{400}\) | B1 | \(\frac{1}{400}\) or 0.0025 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{exactly 1 red}) = 2 \times \frac{1}{20} \times \frac{19}{20} = \frac{19}{200}\) | M1, A1 | M1 for correct equivalent expression \(\frac{1}{20}\times\frac{19}{20}+\frac{19}{20}\times\frac{1}{20}\); A1 \(\frac{19}{200}\) or 0.095 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{2 yellow and 1 green}) = 3 \times \frac{4}{9} \times \frac{5}{8} \times \frac{4}{7} = \frac{10}{21}\) | B1 M1 A1 | B1 for \(3\times\ldots\) or for sum of exactly 3 identical products attempted; M1 for any one product correct; A1 \(\frac{10}{21}\) (allow awrt 0.476 from correct working) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{All beads are yellow}) = \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6}\) | M1 | 1st M1 for \(\frac{5}{9}\times\frac{4}{8}\times\frac{3}{7}\times\frac{2}{6}\) |
| \(P(\text{At least 1 bead is green}) = 1 - P(\text{All beads are yellow})\) \(= 1 - \frac{5}{9}\times\frac{4}{8}\times\frac{3}{7}\times\frac{2}{6} = \frac{121}{126}\) | M1A1 | 2nd M1 use of \(1-p\) where \(p\) is a product of 4 probabilities; A1 \(\frac{121}{126}\) (condone awrt 0.960, must be at least 3sf from correct working) |
## Question 4:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{both blue}) = \frac{1}{20} \times \frac{1}{20} = \frac{1}{400}$ | B1 | $\frac{1}{400}$ or 0.0025 |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{exactly 1 red}) = 2 \times \frac{1}{20} \times \frac{19}{20} = \frac{19}{200}$ | M1, A1 | M1 for correct equivalent expression $\frac{1}{20}\times\frac{19}{20}+\frac{19}{20}\times\frac{1}{20}$; A1 $\frac{19}{200}$ or 0.095 |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{2 yellow and 1 green}) = 3 \times \frac{4}{9} \times \frac{5}{8} \times \frac{4}{7} = \frac{10}{21}$ | B1 M1 A1 | B1 for $3\times\ldots$ or for sum of exactly 3 identical products attempted; M1 for any one product correct; A1 $\frac{10}{21}$ (allow awrt 0.476 from correct working) |
### Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{All beads are yellow}) = \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6}$ | M1 | 1st M1 for $\frac{5}{9}\times\frac{4}{8}\times\frac{3}{7}\times\frac{2}{6}$ |
| $P(\text{At least 1 bead is green}) = 1 - P(\text{All beads are yellow})$ $= 1 - \frac{5}{9}\times\frac{4}{8}\times\frac{3}{7}\times\frac{2}{6} = \frac{121}{126}$ | M1A1 | 2nd M1 use of $1-p$ where $p$ is a product of 4 probabilities; A1 $\frac{121}{126}$ (condone awrt 0.960, must be at least 3sf from correct working) |
---\begin{enumerate}
\item A bag contains 19 red beads and 1 blue bead only.
\end{enumerate}
Linda selects a bead at random from the bag. She notes its colour and replaces the bead in the bag. She then selects a second bead at random from the bag and notes its colour.
Find the probability that\\
(a) both beads selected are blue,\\
(b) exactly one bead selected is red.
In another bag there are 9 beads, 4 of which are green and the rest are yellow.\\
Linda selects 3 beads from this bag at random without replacement.\\
(c) Find the probability that 2 of these beads are yellow and 1 is green.
Linda replaces the 3 beads and then selects another 4 at random without replacement.\\
(d) Find the probability that at least 1 of the beads is green.
\hfill \mbox{\textit{Edexcel S1 2015 Q4 [9]}}