Question
A farmer grows two types of apples—cooking apples and eating apples. The weights of the apples, in grams, can be modelled as normal distributions with the following parameters.
For each type of apple you can assume that 95% of the weights are within two standard deviations of the mean.
(a) Find the percentage of eating apples that have a weight greater than 140 g.
The farmer grows a large number of apples of which 80% are eating apples.
Both types of apples are picked and randomly mixed together in a cleaning machine.
After cleaning, the machine separates out those that have a weight greater than 140 g into a container.
(b) An apple is randomly selected from this container. Find the probability that it is an eating apple. Give your answer in the form $\frac{c}{d}$, where $c, d \in \mathbb{Z}$.
▶️Answer/Explanation
Solution:-
6. (a) 2.5%
(b) P(weight of cooking apples > 140) = 0.5 (50%) (seen anywhere)
recognition of conditional probability in context
$P(\text{eating apple | weight > 140}) = \frac{P(\text{eating apple and weight of eating apple > 140})}{P(\text{weight of apple > 140})}$
$\frac{P(\text{eating apple $\bigcap$ weight of eating apple > 140})}{[P(\text{eating apple $\bigcap$ weight of eating apple > 140}) + P(\text{cooking apple $\bigcap$ weight of cooking apple > 140})]}$
OR
$\frac{P(\text{eating apple}) P(\text{weight of eating apple > 140})}{[P(\text{eating apple}) P(\text{weight of eating apple >140}) + P(\text{cooking apple}) P(\text{weight of cooking apple | 140})]}$
= $\frac{0.8 \times 0.025}{0.8 \times 0.025 + 0.2 \times 0.5} = \frac{80 \times 2.5}{80 \times 2.5 + 20 \times 50} = \frac{200}{1200} = \frac{1}{6}$
Question
At a nursing college, 80 % of incoming students are female. College records show that 70 % of the incoming females graduate and 90 % of the incoming males graduate. A student who graduates is selected at random. Find the probability that the student is male, giving your answer as a fraction in its lowest terms.
▶️Answer/Explanation
Markscheme
\({\text{P }}M|G = \frac{{{\text{P}}(M \cap G)}}{{{\text{P}}(G)}}\) (M1)
\( = \frac{{0.2 \times 0.9}}{{0.2 \times 0.9 + 0.8 \times 0.7}}\) M1A1A1
\( = \frac{{0.18}}{{0.74}}\)
\( = \frac{9}{{37}}\) A1
[5 marks]
Examiners report
Most candidates answered this question successfully. Some made arithmetic errors.
Question
Jenny goes to school by bus every day. When it is not raining, the probability that the bus is late is \(\frac{3}{{20}}\). When it is raining, the probability that the bus is late is \(\frac{7}{{20}}\). The probability that it rains on a particular day is \(\frac{9}{{20}}\). On one particular day the bus is late. Find the probability that it is not raining on that day.
▶️Answer/Explanation
Markscheme
(A1)
\({\text{P}}(R’ \cap L) = \frac{{11}}{{20}} \times \frac{3}{{20}}\) A1
\({\text{P}}(L) = \frac{9}{{20}} \times \frac{7}{{20}} + \frac{{11}}{{20}} \times \frac{3}{{20}}\) A1
\({\text{P}}(R’|L) = \frac{{{\text{P}}(R’ \cap L)}}{{{\text{P}}(L)}}\) (M1)
\( = \frac{{33}}{{96}}{\text{ }}\left( { = \frac{{11}}{{32}}} \right)\) A1
[5 marks]
Examiners report
This question was generally well answered with candidates who drew a tree diagram being the most successful.