The table below shows the number of male and female students studying Chemistry, Physics and Biology at a private school.

	Chemistry	Physics	Biology
Male	200	130	70
Female	250	300	50

One of the students is chosen at random. Events $C$, $B$ and $M$ are defined as follows: $C$: The student chosen is studying Chemistry. $B$: The student chosen is studying Biology. $M$: The student chosen is a male.

1. Determine the total number of male students

$\text{Total male students } = 200 + 130 + 70 = 400$

2. Calculate the conditional probability P(C|M)

$\text{P}(C|M) = \frac{n(C \cap M)}{n(M)} = \frac{200}{400} = \frac{1}{2}$

1. Calculate the total number of students in the school

$\text{Total students} = 200 + 130 + 70 + 250 + 300 + 50 = 1000$

2. Find the number of students who are male or studying Chemistry, and calculate the probability

The number of students who are male or studying Chemistry is the sum of all males plus the females studying Chemistry:
$n(M \cup C) = 400 + 250 = 650$
$\text{P}(M \cup C) = \frac{650}{1000} = \frac{13}{20}$

1. Determine the number of students who are both female ($M'$) and not studying Biology ($B'$)

$M'$ represents female students and $B'$ represents students who do not study Biology (i.e. study Chemistry or Physics).

$\text{Number of students in } M' \cap B' = 250 \text{ (Female, Chem)} + 300 \text{ (Female, Physics)} = 550$

2. Calculate the probability

$\text{Total number of students} = 200+130+70+250+300+50 = 1000$

$\text{P}(M' \cap B') = \frac{550}{1000} = \frac{11}{20} \text{ (or } 0.55)$

1. Calculate the individual probabilities of events $C$, $M$, and their intersection $C \cap M$.

Total number of students $= 400 \text{ (males)} + 600 \text{ (females)} = 1000$.
Total studying Chemistry $= 200 \text{ (males)} + 250 \text{ (females)} = 450$.

$\text{P}(C) = \frac{450}{1000} = 0.45$
$\text{P}(M) = \frac{400}{1000} = 0.4$
$\text{P}(C \cap M) = \frac{200}{1000} = 0.2$

2. Test the condition for independence and conclude.

$\text{P}(C) \times \text{P}(M) = 0.45 \times 0.4 = 0.18$

Since $\text{P}(C \cap M) \neq \text{P}(C) \times \text{P}(M)$ (as $0.2 \neq 0.18$), the events $C$ and $M$ are not independent.

1. Calculate the total number of students taking each subject.

$n(\text{Chemistry}) = 200 + 250 = 450$
$n(\text{Physics}) = 130 + 300 = 430$
$n(\text{Biology}) = 70 + 50 = 120$

2. Calculate the number of international students in each subject and find the overall total.

$n(\text{Int. Chemistry}) = 0.20 \times 450 = 90$
$n(\text{Int. Physics}) = 0.30 \times 430 = 129$
$n(\text{Int. Biology}) = 0.05 \times 120 = 6$
$\text{Total Int. Students} = 90 + 129 + 6 = 225$

3. Calculate the conditional probability that the student studies Chemistry given they are an international student.

$\text{P}(\text{Chemistry} \mid \text{Int.}) = \frac{n(\text{Int. Chemistry})}{n(\text{Total Int. Students})}$
$= \frac{90}{225}$
$= \frac{2}{5}$

1. Find the number of international students studying Physics and the total number of students.

Total number of students $= 200 + 250 + 130 + 300 + 70 + 50 = 1000$
Total number of Physics students $= 130 + 300 = 430$
Number of international Physics students $= 0.30 \times 430 = 129$

2. Calculate the probability of choosing exactly one international Physics student out of three.

Number of students who are NOT international Physics students $= 1000 - 129 = 871$

The probability of choosing exactly one such student in three draws (without replacement) requires accounting for the 3 different possible orders (e.g., Yes-No-No, No-Yes-No, No-No-Yes):
$\text{P(exactly 1)} = 3 \times \left( \frac{129}{1000} \times \frac{871}{999} \times \frac{870}{998} \right)$
$= \frac{48876165}{166167000} \approx 0.29414 \approx 0.294 \text{ (to 3 s.f.)}$