Pre-U Pre-U 9794/1 Specimen — Question 15 12 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
SessionSpecimen
Marks12
TopicConditional Probability
TypeConditional with three or more stages
DifficultyStandard +0.8 This is a multi-stage conditional probability problem requiring careful tracking of changing probabilities across three stages, construction of a tree diagram with 8 outcomes, and calculation of conditional probability in part (iv). While the individual calculations are straightforward, the problem requires systematic organization and the final part demands understanding of conditional probability with a non-trivial conditioning event, placing it moderately above average difficulty.
Spec2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables

15 In order to be accepted on a university course, a student needs to pass three exams.
The probability that the student passes the first exam is \(\frac { 3 } { 4 }\).
For each of the second and third exams, the probability of passing the exam is
  • the same as the probability of passing the preceding exam if the student passed the preceding exam,
  • half of the probability of passing the preceding exam if the student failed the preceding exam.
    1. Draw a tree diagram to represent the above information.
    2. Find the probability that the student passes all three exams.
    3. Find the probability that the student passes at least two of the exams.
    4. Find the probability that the student passes the third exam given that exactly two of the three exams are passed.

(i) [Tree diagram with correct structure] M1
Top half correct A1
Bottom half correct A1 [3]
(ii) Multiply along the correct path M1
\(\left(\frac{3}{4}\right)^3 = \frac{27}{64} = 0.422\) A1 [2]
(iii) Attempt *EITHER*: \(P(PFP) + P(FPP) + P(PPF)\)
*OR*: \(1 - [P(FFF) + P(FPF) + P(PFF)]\) M1
\(= \frac{27}{64} + \frac{9}{128} + \frac{9}{256} + \frac{9}{64} = \frac{171}{256} = 0.668\) A1 [2]
(iv) \(P(PFP) + P(FPP)\) M1
\(= \frac{9}{128} + \frac{9}{256} = \frac{27}{256} = 0.1055\) A1
\(P(PFP) + P(FPP) + P(PPF) = \frac{9}{128} + \frac{9}{256} + \frac{9}{64} = \frac{63}{256} = 0.246\) B1\(\checkmark\)
Divide correct combinations M1
\(= \frac{3}{7} = 0.429\) A1 [5]
**(i)** [Tree diagram with correct structure] M1

Top half correct A1

Bottom half correct A1 **[3]**

**(ii)** Multiply along the correct path M1

$\left(\frac{3}{4}\right)^3 = \frac{27}{64} = 0.422$ A1 **[2]**

**(iii)** Attempt *EITHER*: $P(PFP) + P(FPP) + P(PPF)$

*OR*: $1 - [P(FFF) + P(FPF) + P(PFF)]$ M1

$= \frac{27}{64} + \frac{9}{128} + \frac{9}{256} + \frac{9}{64} = \frac{171}{256} = 0.668$ A1 **[2]**

**(iv)** $P(PFP) + P(FPP)$ M1

$= \frac{9}{128} + \frac{9}{256} = \frac{27}{256} = 0.1055$ A1

$P(PFP) + P(FPP) + P(PPF) = \frac{9}{128} + \frac{9}{256} + \frac{9}{64} = \frac{63}{256} = 0.246$ B1$\checkmark$

Divide correct combinations M1

$= \frac{3}{7} = 0.429$ A1 **[5]**
15 In order to be accepted on a university course, a student needs to pass three exams.\\
The probability that the student passes the first exam is $\frac { 3 } { 4 }$.\\
For each of the second and third exams, the probability of passing the exam is

\begin{itemize}
  \item the same as the probability of passing the preceding exam if the student passed the preceding exam,
  \item half of the probability of passing the preceding exam if the student failed the preceding exam.\\
(i) Draw a tree diagram to represent the above information.\\
(ii) Find the probability that the student passes all three exams.\\
(iii) Find the probability that the student passes at least two of the exams.\\
(iv) Find the probability that the student passes the third exam given that exactly two of the three exams are passed.
\end{itemize}

\hfill \mbox{\textit{Pre-U Pre-U 9794/1  Q15 [12]}}