Edexcel S1 — Question 6 12 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConditional Probability
TypeSampling without replacement from bags/boxes
DifficultyModerate -0.3 This is a straightforward S1 conditional probability question involving sampling without replacement. Part (a) is a 'show that' calculation requiring P(M₁∩M₂) = (4/10)×(3/9) = 2/15. Part (b) requires enumerating cases (FFF, FFM, FMF, MFF) with basic probability calculations. Part (c) is direct application of conditional probability P(F₂∩F₃|M₁) = (6/9)×(5/8). All parts use standard techniques with no novel insight required, making it slightly easier than a typical A-level question due to its routine nature and clear structure.
Spec2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles

6. At the start of a gameshow there are 10 contestants of which 6 are female. In each round of the game, one contestant is eliminated. All of the contestants have the same chance of progressing to the next round each time.
  1. Show that the probability that the first two contestants to be eliminated are both male is \(\frac { 2 } { 15 }\).
  2. Find the probability that more females than males are eliminated in the first three rounds of the game.
  3. Given that the first contestant to be eliminated is male, find the probability that the next two contestants to be eliminated are both female.
    (3 marks)

AnswerMarks Guidance
(a) \(\frac{7}{10} \times \frac{3}{9} = \frac{7}{30}\)M2 A1
(b) \(P(\text{more F}) = P(3F) + P(2F)\)M1
\(= (\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}) + (3 \times \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}) = \frac{2}{3}\)M3 A2
(c) After M goes, left with 6F and 3MM1
\(P(\text{next 2 F}) = \frac{6}{9} \times \frac{5}{8} = \frac{5}{12}\)M1 A1 (12 marks)
(a) $\frac{7}{10} \times \frac{3}{9} = \frac{7}{30}$ | M2 A1 |

(b) $P(\text{more F}) = P(3F) + P(2F)$ | M1 |

$= (\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}) + (3 \times \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}) = \frac{2}{3}$ | M3 A2 |

(c) After M goes, left with 6F and 3M | M1 |

$P(\text{next 2 F}) = \frac{6}{9} \times \frac{5}{8} = \frac{5}{12}$ | M1 A1 | (12 marks)

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6. At the start of a gameshow there are 10 contestants of which 6 are female. In each round of the game, one contestant is eliminated. All of the contestants have the same chance of progressing to the next round each time.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that the first two contestants to be eliminated are both male is $\frac { 2 } { 15 }$.
\item Find the probability that more females than males are eliminated in the first three rounds of the game.
\item Given that the first contestant to be eliminated is male, find the probability that the next two contestants to be eliminated are both female.\\
(3 marks)
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1  Q6 [12]}}