Pre-U Pre-U 9794/3 2015 June — Question 2 6 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2015
SessionJune
Marks6
TopicGeometric Distribution
TypeGeometric with multiple success milestones
DifficultyStandard +0.3 This is a straightforward application of geometric distribution concepts with standard probability calculations. Part (a) requires basic counting principles (5!/5^5), part (b)(i) is direct recognition of Geo(1/5), and parts (b)(ii-iii) use standard formulas for expectation and P(X≥3). While it involves multiple parts, each step follows textbook methods without requiring novel insight or complex multi-stage reasoning.
Spec2.03a Mutually exclusive and independent events5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2

2 Jill is collecting picture cards given away in packets of a particular brand of breakfast cereal. There are five different cards in the complete set. Each packet contains one card which is equally likely to be any of the five cards in the set.
  1. Find the probability that Jill has a complete set of cards from the first five packets that she buys.
  2. At some point Jill needs just one more card to complete the set. Let \(X\) be the random variable that represents the number of additional packets that Jill will need to buy in order to complete the set.
    1. Write down the distribution of \(X\).
    2. State the expected number of additional packets that Jill will need to buy.
    3. Find the probability that Jill will need to buy at least 3 additional packets in order to complete the set.

Question 2 [8 marks]
(a) \(\frac{5!}{5^5} = \frac{120}{3125} = \frac{24}{625} = 0.0384\)
— M1: Product of 5 probabilities, at least 4 correct. A1: c.a.o. Either fraction or decimal. [2]
(b)(i) \(X \sim \text{Geo}\!\left(\frac{1}{5}\right)\)
— B1: Must give parameter as well as name. [1]
(b)(ii) \(E(X) = 5\)
— B1: Allow \(\frac{1}{\textit{their } p}\) from (ii). [1]
(b)(iii) \(P(X \geqslant 3) = \left(\frac{4}{5}\right)^2 = \frac{16}{25} = 0.64\)
— M1: Attempt \(P(X > 3)\) or equivalent methods. A1: c.a.o. Either fraction or decimal. [2]
Total: [6]
**Question 2** [8 marks]

**(a)** $\frac{5!}{5^5} = \frac{120}{3125} = \frac{24}{625} = 0.0384$
— M1: Product of 5 probabilities, at least 4 correct. A1: c.a.o. Either fraction or decimal. **[2]**

**(b)(i)** $X \sim \text{Geo}\!\left(\frac{1}{5}\right)$
— B1: Must give parameter as well as name. **[1]**

**(b)(ii)** $E(X) = 5$
— B1: Allow $\frac{1}{\textit{their } p}$ from **(ii)**. **[1]**

**(b)(iii)** $P(X \geqslant 3) = \left(\frac{4}{5}\right)^2 = \frac{16}{25} = 0.64$
— M1: Attempt $P(X > 3)$ or equivalent methods. A1: c.a.o. Either fraction or decimal. **[2]**

**Total: [6]**
2 Jill is collecting picture cards given away in packets of a particular brand of breakfast cereal. There are five different cards in the complete set. Each packet contains one card which is equally likely to be any of the five cards in the set.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Jill has a complete set of cards from the first five packets that she buys.
\item At some point Jill needs just one more card to complete the set. Let $X$ be the random variable that represents the number of additional packets that Jill will need to buy in order to complete the set.
\begin{enumerate}[label=(\roman*)]
\item Write down the distribution of $X$.
\item State the expected number of additional packets that Jill will need to buy.
\item Find the probability that Jill will need to buy at least 3 additional packets in order to complete the set.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2015 Q2 [6]}}