Pre-U Pre-U 9794/3 2015 June — Question 5 12 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2015
SessionJune
Marks12
TopicBinomial Distribution
TypeTwo-stage binomial problems
DifficultyStandard +0.3 This is a straightforward two-stage binomial problem requiring identification of B(3, 0.7), calculation of standard binomial probabilities, recognition that 'at most 1 seedling' creates a modified distribution, and a final compound probability calculation. All steps are routine applications of binomial distribution with clear structure and no novel insight required.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities

5 A garden centre grows a particular variety of plant for sale. They sow 3 seeds in each pot and there are 6 pots in a tray. The probability that a seed germinates is 0.7 , independently of any other seeds.
  1. State the probability distribution of the number of seeds in a pot that germinate.
  2. Find the probability that, in a randomly chosen pot,
    1. exactly 2 seeds germinate,
    2. at least 1 seed germinates. After the seeds have germinated and become seedlings, some are removed (and discarded) so that there remains at most 1 seedling per pot.
    3. Write out the probability distribution of the number of seedlings per pot that remain.
    4. Find the probability that there is a seedling in every one of the 6 pots in a randomly chosen tray.

Question 5
(i) \((X \sim)\ \text{Bin}(3,\ 0.7)\)
— B2: All 3 elements present and correct. Allow B1 for only 1 error/omission. [2]
(ii)(a) \(P(X = 2) = 3 \times 0.7^2 \times 0.3\)
\(= 0.441\)
— M1: \(^3C_2 \times \ldots\) M1: \(\ldots p^2 \times q\). A1: c.a.o. [3]
(ii)(b) \(P(X \geqslant 1) = 1 - 0.3^3\)
\(= 0.973\)
— M1: Or by summing \(P(1) \ldots P(3)\). A1: c.a.o. [2]
(iii)
AnswerMarks Guidance
\(x\)0 1
\(P(X = x)\)0.027 0.973
— B1: Values in top row. B1: \(P(X=1) = \textit{their } \textbf{(ii)(b)}\). B1: \(P(X=0) = 1 - \textit{their } \textbf{(ii)(b)}\). \(\text{Bin}(1,\ 0.973)\) earns full marks. [3]
(iv) \(P(\text{All contain a seedling}) = 0.973^6\)
\(= 0.84854\ldots \approx 0.849\)
— M1: Ft *their* \(P(X=1)\). A1: c.a.o. [2]
**Question 5**

**(i)** $(X \sim)\ \text{Bin}(3,\ 0.7)$
— B2: All 3 elements present and correct. Allow B1 for only 1 error/omission. **[2]**

**(ii)(a)** $P(X = 2) = 3 \times 0.7^2 \times 0.3$
$= 0.441$
— M1: $^3C_2 \times \ldots$ M1: $\ldots p^2 \times q$. A1: c.a.o. **[3]**

**(ii)(b)** $P(X \geqslant 1) = 1 - 0.3^3$
$= 0.973$
— M1: Or by summing $P(1) \ldots P(3)$. A1: c.a.o. **[2]**

**(iii)**

| $x$ | 0 | 1 |
|---|---|---|
| $P(X = x)$ | 0.027 | 0.973 |

— B1: Values in top row. B1: $P(X=1) = \textit{their } \textbf{(ii)(b)}$. B1: $P(X=0) = 1 - \textit{their } \textbf{(ii)(b)}$. $\text{Bin}(1,\ 0.973)$ earns full marks. **[3]**

**(iv)** $P(\text{All contain a seedling}) = 0.973^6$
$= 0.84854\ldots \approx 0.849$
— M1: Ft *their* $P(X=1)$. A1: c.a.o. **[2]**
5 A garden centre grows a particular variety of plant for sale. They sow 3 seeds in each pot and there are 6 pots in a tray. The probability that a seed germinates is 0.7 , independently of any other seeds.\\
(i) State the probability distribution of the number of seeds in a pot that germinate.\\
(ii) Find the probability that, in a randomly chosen pot,
\begin{enumerate}[label=(\alph*)]
\item exactly 2 seeds germinate,
\item at least 1 seed germinates.

After the seeds have germinated and become seedlings, some are removed (and discarded) so that there remains at most 1 seedling per pot.\\
(iii) Write out the probability distribution of the number of seedlings per pot that remain.\\
(iv) Find the probability that there is a seedling in every one of the 6 pots in a randomly chosen tray.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2015 Q5 [12]}}