| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Multiple independent trials |
| Difficulty | Moderate -0.8 This is a straightforward AS-level probability question involving independent trials with given probabilities. Part (a) uses the complement rule (1 - P(no bananas)), part (b) requires summing three simple cases, and part (c) is a direct binomial coefficient calculation. All parts are standard textbook exercises requiring only routine application of basic probability rules with no problem-solving insight needed. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| 14 | apple twice and cake twice |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{at least 1 banana}) = 1 - P(\text{no bananas}) = 1 - 0.65^4\) | M1 | Uses \(1 - p^4\) for \(0 < p < 1\); or uses \(P(1)+P(2)+P(3)+P(4)\); or uses \(P(x \geq 1)\) using \(B(4, 0.35)\) |
| \(= 0.82149\) | A1 | AWFW 0.821 to 0.822; accept 0.82 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.2^4 + 0.35^4 + 0.45^4\) | M1 | Calculates the correct three terms, not necessarily evaluated and not necessarily added |
| \(= 0.0576125\) | A1 | AWRT 0.058 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{apple twice and cake twice})\); uses \(0.2^2\) or \(0.45^2\) anywhere | B1 | Or fraction equivalents |
| \(= 6 \times 0.2^2 \times 0.45^2\) | M1 | Uses \((k \times) \ 0.2^2 \times 0.45^2 (\times 0.35^0)\); or fraction equivalents and no other terms; PI by correct answer |
| \(= 0.0486\) | A1 | CAO |
## Question 14(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{at least 1 banana}) = 1 - P(\text{no bananas}) = 1 - 0.65^4$ | M1 | Uses $1 - p^4$ for $0 < p < 1$; or uses $P(1)+P(2)+P(3)+P(4)$; or uses $P(x \geq 1)$ using $B(4, 0.35)$ |
| $= 0.82149$ | A1 | AWFW 0.821 to 0.822; accept 0.82 |
## Question 14(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.2^4 + 0.35^4 + 0.45^4$ | M1 | Calculates the correct three terms, not necessarily evaluated and not necessarily added |
| $= 0.0576125$ | A1 | AWRT 0.058 |
## Question 14(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{apple twice and cake twice})$; uses $0.2^2$ or $0.45^2$ anywhere | B1 | Or fraction equivalents |
| $= 6 \times 0.2^2 \times 0.45^2$ | M1 | Uses $(k \times) \ 0.2^2 \times 0.45^2 (\times 0.35^0)$; or fraction equivalents and no other terms; PI by correct answer |
| $= 0.0486$ | A1 | CAO |
---14 Yingtai visits her local gym regularly.
After each visit she chooses one item to eat from the gym's cafe.\\
This could be an apple, a banana or a piece of cake.\\
She chooses the item independently each time.\\
The probability that Yingtai chooses each of these items on any visit is given by:
$$\begin{aligned}
\mathrm { P } ( \text { Apple } ) & = 0.2 \\
\mathrm { P } ( \text { Banana } ) & = 0.35 \\
\mathrm { P } ( \text { Cake } ) & = 0.45
\end{aligned}$$
For any four randomly selected visits to the gym, find the probability that Yingtai chose:
14
\begin{enumerate}[label=(\alph*)]
\item at least one banana.\\[0pt]
[2 marks]\\
14
\item the same item each time.\\
\begin{center}
\begin{tabular}{|l|l|}
\hline
14
\item & apple twice and cake twice \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2022 Q14 [7]}}