Edexcel S1 — Question 7 16 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeSampling without replacement until success
DifficultyStandard +0.3 This is a straightforward S1 question on sampling without replacement with clear structure and standard calculations. Part (a) guides students through the method, parts (b-c) are routine probability distribution work, and parts (d-e) apply expected value in a simple context. The calculations are manageable and the question type is common in textbooks, making it slightly easier than average.
Spec5.01a Permutations and combinations: evaluate probabilities5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables

7. A bag contains 4 red and 2 blue balls, all of the same size. A ball is selected at random and removed from the bag. This is repeated until a blue ball is pulled out of the bag. The random variable \(B\) is the number of balls that have been removed from the bag.
  1. Show that \(\mathrm { P } ( B = 2 ) = \frac { 4 } { 15 }\).
  2. Find the probability distribution of \(B\).
  3. Find \(\mathrm { E } ( B )\). The bag and the same 6 balls are used in a game at a funfair. One ball is removed from the bag at a time and a contestant wins 50 pence if one of the first two balls picked out is blue.
  4. What are the expected winnings from playing this game once? For \(\pounds 1\), a contestant gets to play the game three times.
  5. What is the expected profit or loss from the three games?

AnswerMarks Guidance
(a) \(\frac{4}{6} \times \frac{2}{5} = \frac{4}{15}\)M1 A1
(b) Same method, givingM2 A2
\(b\)1 2
\(P(B = b)\)\(\frac{1}{3}\) \(\frac{4}{15}\)
(c) \(\sum bP(b) = \frac{1}{15}(5 + 8 + 9 + 8 + 5) = \frac{35}{15} = \frac{7}{3}\)M2 A1
(d) \(P(\text{winning}) = \frac{1}{3} + \frac{4}{15} = \frac{3}{5}\)M1 A1
expected winnings \(= \frac{3}{5} \times 50 = 30\) penceM1 A1
(e) \((3 \times 30) - 100 = -10 \therefore\) 10 pence lossM2 A1 (16 marks total)
Total: 75 marks
(a) $\frac{4}{6} \times \frac{2}{5} = \frac{4}{15}$ | M1 A1 |

(b) Same method, giving | M2 A2 |

| $b$ | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| $P(B = b)$ | $\frac{1}{3}$ | $\frac{4}{15}$ | $\frac{1}{5}$ | $\frac{2}{15}$ | $\frac{1}{15}$ |

(c) $\sum bP(b) = \frac{1}{15}(5 + 8 + 9 + 8 + 5) = \frac{35}{15} = \frac{7}{3}$ | M2 A1 |

(d) $P(\text{winning}) = \frac{1}{3} + \frac{4}{15} = \frac{3}{5}$ | M1 A1 |

expected winnings $= \frac{3}{5} \times 50 = 30$ pence | M1 A1 |

(e) $(3 \times 30) - 100 = -10 \therefore$ 10 pence loss | M2 A1 | (16 marks total)

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**Total: 75 marks**
7. A bag contains 4 red and 2 blue balls, all of the same size. A ball is selected at random and removed from the bag. This is repeated until a blue ball is pulled out of the bag.

The random variable $B$ is the number of balls that have been removed from the bag.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( B = 2 ) = \frac { 4 } { 15 }$.
\item Find the probability distribution of $B$.
\item Find $\mathrm { E } ( B )$.

The bag and the same 6 balls are used in a game at a funfair. One ball is removed from the bag at a time and a contestant wins 50 pence if one of the first two balls picked out is blue.
\item What are the expected winnings from playing this game once?

For $\pounds 1$, a contestant gets to play the game three times.
\item What is the expected profit or loss from the three games?
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1  Q7 [16]}}